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FUNDAMENTALS 
)F HIGH SCHOOL 
MATHEMATICS 

RUG G -CLARK 




I 

RUGG-CLARK MATHEMATICS TEXTS 

By Harold O. Rugg and John R. Clark 

Fundamentals of High School 
Mathematics 

A Textbook Designed to Follow Arithmetic 



BY 

HAROLD O. RUGG 

Associate Professor of Education 
University of Chicago 

AND 

JOHN R. CLARK 

Head of Department of Mathematics 
Francis W. Parker School, Chicago 



Experimental Edition, distributed at cost for experimental 
purposes only. NOT for general commercial distribution 



1918 

WORLD BOOK COMPANY 

YONKERS-ON-HUDSON, NEW YORK 
2126 PRAIRIE AVENUE, CHICAGO 



WORLD BOOK COMPANY 

THE HOUSE OF APPLIED KNOWLEDGE 

Established, 1905, by Caspar W. Hodgson 

yonkers-on-hudson, new york 
2126 Prairie Avenue, Chicago 



The purpose of this house is to publish 
books that apply the world's knowledge to 
the world's needs. Public-school texts in 
mathematics should be designed in terms of 
careful studies of social needs as well as in 
terms of methods by which pupils learn. 
This book on general mathematics, the first 
of the Rugg-Clark Mathematics Texts, is the 
outgrowth of years of scientific investigation 
and of experimental teaching. It will be 
followed by texts for Junior High Schools 
and for Elementary Schools, each of which 
will be published only after years of scien- 
tific investigation and experimentation. 



SEP -3 1918 



©CI.A503268 

rcmt: fhsm-i 



Copyright, 1918, by World Book Company 
All rights reserved 









u.^/lr^ OJMMJ& QmulaJUm^j 



FOREWORD 

THE CONSTRUCTION OF COURSES OF STUDY IN HIGH SCHOOL 
MATHEMATICS 

The one-year course of study represented by this textbook is 
the outcome of five years of critical investigation of high-school 
mathematics. A recent monograph published by the writers * 
presents in detail the evidence developed by these years of in- 
vestigation. It reveals striking weaknesses in the content and 
teaching of first-year algebra. It shows that the achievement 
of more than 22.000 pupils on carefully designed standardized 
tests is very unsatisfactory ; that the present course of study is 
not organized primarily to provide an opportunity for training in 
problem-solving ; and that much of the material included in the 
traditional course will never be used in actual life situations. 
The evidence is clear, therefore, that the present course of study 
in first-year algebra must be completely reconstructed. 

The study of the historical development of the present first- 
year high school algebra shows that it has come down from its 
former position in the upper college curriculum with but two 
slight modifications of content and presentation. One has been 
revealed in the increased emphasis on the formal and manipu- 
lative aspects : the second in the slight improvements which 
have been made in methods of presentation through the use of 
more concrete devices. On the whole, however, the course has 
retained that emphasis upon rigorous, logical organization which 
characterizes the mathematical thinking of " mathematicians." 

The need is evident, therefore, for a course of study which 
will avoid the two fundamental weaknesses pointed out above ; 
namely, first, a course which will eliminate formal non-essentials, 
basing such elimination on scientific investigation ; second, one 
which will tie together, in a psychologically and sequentially 
worked-out scheme, the fundamental mathematical notions and 
tool operations which are needed by adults to facilitate quanti- 

1 Scientific Method in the Reco?istmction of Xinth- Grade Mathematics. 
University of Chicago Press. 1918. 

iii 



iv Foreword 

tative thinking. The course of study presented in this text is 
therefore unique in these particulars. 

PRINCIPLES OF TEXTBOOK DESIGN 

Three important criteria have been used to determine the 
content and the organization of this experimental course of 
study : first, subject matter must be organized in terms of a real 
psychological analysis of " learning " in mathematics ; second, 
those mathematical notions and devices which have the widest 
application must be emphasized roughly in proportion to their 
frequency of use ; thus the material has been selected in terms 
of the social criterion ; third, the course must be organized to 
provide the pupil with the maximum opportunity to do genuine 
thinking, real problem-solving, rather than to emphasize the 
drill or manipulative aspects which now commonly require most 
of the pupil's time. 

The first criterion. To organize a course of study in terms of 
a real psychology of " learning" in mathematics shows us that 
new meanings, new concepts which are to be learned, must be 
acquired by the pupil in the same natural way in which human 
beings learn and acquire new meanings outside of a textbook. 
To be more specific, this means that the exact logical sequence 
of the mature mathematician, which begins with abstract defini- 
tions and statements of general principles and then proceeds to 
their applications and specific uses, must give way to a directly 
psychological method. " Definitions " and general principles 
must grow out of the pupil's concrete experience. He must 
begin with details, particulars, concrete elements, and finally 
arrive at a generalization or the statement of a general principle 
(seldom, however, at " definitions " in first-ye^r algebra). Con- 
trast the introductory lesson in algebra which explains that 
"Algebra is the science of general numbers" with the one 
which begins, " It saves time to use abbreviations or letters to 
represent numbers ; for example, you have used C to stand for 
ioo, M for iooo,etc." The authors have attempted to visualize 



Foreword v 

the pupil's mental background \w. writing each page of the exposi- 
tion of this text and in the selection and statement of problem 
material. This represents at least a first step toward the design- 
ing of textbooks on this psychological criterion. 

The second criterion. The authors' investigations of the math- 
ematical experience which pupils need to have to prepare them 
to meet adequately later quantitative situations have resulted in 
the elimination of a great deal of the content of the traditional 
book. The time usually given to the operations on polynomials, 
special products, factoring, complicated fractions, etc., can have 
no justification in terms of this criterioji of social wo?'tk or utility. 
The application of this criterion demands a complete recognition 
of the graphic method of representing number as one of the 
three methods around which a course should be constructed, and 
that the formula, the equation, the properties of the more im- 
portant space forms, and the principle of dependence or func- 
tionality should form the basic material of the course. Likewise 
three chapters of the text are devoted to a non-demonstrative 
study of the triangle. Pupils are shown that mathematics sup- 
plies notions and devices which people need to master in order 
to solve many practical problems. The use of scale drawings, 
angular measurement, the principle of similarity, and the simple 
trigonometric functions of a right triangle have infinitely more 
value either (i) from the standpoint of their use in other situa- 
tions, (2) from their appropriateness and adaptability to the 
child's interests and abilities, or (3) from the criterion of think- 
ing value, than does the excessive formalism and manipulation 
of symbols which they supplant in this course. To summarize, 
the authors have been concerned to incorporate here the most 
important mathematical notions which investigation shows that 
all pupils ought to know. 

The third criterion. Experimental investigation of the possi- 
bility of developing powers of generalization shows that such are 
to be developed only by so complete an organization of courses 
as will provide a maximum of opportunity for problem-solving, 



vi Foreword 

for reflective thinking, for calling into play mental processes of 
analysis, comparison, and recognition of relations between the 
parts of problems. Unfortunately, algebra courses have been 
deprived of most of their " training " value. An emphasis upon 
formalism, drill, the routine practice in manipulation of mean- 
ingless symbols, and lack of genuine motive are typical examples 
of the way in which we have hampered teachers in the develop- 
ment of problem-solving abilities. The general practice of 
devoting 80 per cent of the problem-material to these formal 
drill examples, leaving only 20 per cent for the verbal problem, — 
which of all types provides most completely opportunity for 
"thinking," — has been radically modified in the course sub- 
mitted in this book. The entire course has been organized 
around a central core of " problem-solving. " Even the purely 
formal materials themselves have been so organized, wherever 
possible, as to provide an opportunity for real thinking and not 
mere habit formation. 

THE PRIMARY FUNCTION OF MATHEMATICAL INSTRUCTION 

The writers' thesis in constructing this course of study is this : 
The central element in human thinking is seeing relationships 
clearly. In the same way the primary function of a high-school 
course in mathematics is to give ability to recognize verbally stated 
relationships between magnitudes, to represent such relationships 
economically by means of symbols, and to determine such relation- 
ships. To carry out this aim the course of study, therefore, 
should be organized in such a way as to develop ability in the in- 
telligent use of the equation, the formula, methods of graphic repre- 
sentation, and the properties of the more important space forms in 
the expression and deter??iination of relationships. 

Thus we may summarize the chief characteristics of this text, 
which has been organized on this fundamental aim, as follows : 

(1) A marked decrease in the emphasis upon formal manipu- 
lation. The whole course is aimed at providing an opportunity 
for problem-solving. 



Foreword vii 

(2) Three methods of representing number are recognized : 
the tabular method, the graphic method, and the equational or 
formula method. Thus graphic representation is an integral 
part of the course and is not treated as an isolated operation. 

(3) A vast amount of useless material has been omitted, — 
for instance, addition, subtraction, multiplication, and division of 
polynomials ; the more complicated work with fractions, all but 
the simpler work with radicals, etc. 

(4) All material shown by investigation to have social utility, 
and which is omitted from current courses in mathematics, has 
been included. For example : construction and evaluation of 
formulas, emphasis on " evaluation " as a most important opera- 
tion, trigonometric functions of the right triangle, a very com- 
plete study of variation or dependence, etc. 

(5) The exposition of the text develops so gradually in accord- 
ance with the writers' discoveries concerning " learning " that 
the average pupil can read any part of the discussion and then 
solve the problems unaided. Thus the text develops in accord- 
ance with the natural development of the pupil's method rather 
than that of the highly trained, logical mathematician. This is 
the outcome of an original analysis of the psychology of mathe- 
matics. 

(6) The equation is emphasized throughout as the primary 
formal operation of the course, — not as an end in itself, but 
because it is the essential tool for stating and determining quan- 
titative relationships. 

THE TEACHER'S TESTS FOR A TEXTBOOK 

The authors would propose to teachers as tests of the value 
of material contained in this or in any other textbook the three 
following principles : (1) Is the type of subject matter presented 
here or in any other textbook organized in terms of the way 
children naturally learn ; that is, has the psychological criterion 
been kept constantly and adequately in mind? (2) Is the pupil 
who takes this or any other course prompted to do real and 



viii Foreword 

genuine thinking ? Does he have ample opportunity for practice 
in "problem-solving"? Is the subject matter of the course 
organized primarily around a core of problem-solving situations ? 
1 )oes the kind of subject matter presented in this or in any 
other book sufficiently justify itself from the point of view of its 
use or importance either in later mathematics courses, in other 
school subjects, or in situations outside the school ? • 

The application of these criteria in the construction and 
selection of school textbooks will go far to bring about the type 
of reconstruction for which the writers' investigations show there 
is a real demand. 

Harold O. Rugg 
Chicago, Illinois John R. Clark 

July 25, 1918 



CONTENTS 

CHAPTER PAGE 

I. A New Way to Represent Numbers . . i 
II. How to Construct and Evaluate Formulas. 17 

III. How to Use the Equation .... 29 

IV. How to Represent the Relationship between 

Quantities w t hich Change Together . . 45 

V. How to Find Unknown Distances by Means 

of Scale Drawings : The First Method . 64 

VI. A Second Method of Finding Unknown Dis- 
tances : The Use of Similar Triangles . 86 

VII. How to Find Unknowns by Means of the Ra- 
tios of the Sides of the Right Triangle . 97 

VIII. How to Show the Way in which One Varying 

Quantity Depends upon Another. . .119 

IX. The Use of Positiye and Negative Numbers 130 

X. The Complete Solution of the Simple Equa- 
tion . . . . . . . -152 

XI. How to Solve Equations which Contain Two 

Unknowns . . . . . . .183 

XII. How to Solve Equations with Two Unknowns 

{Continued) . . . . . . .198 

XIII. How to Find Products and Factors . .210 

XIV. How to Solve Equations of the Second De- 

gree ........ 232 

XV. Further Use of the Right Triangle : How to 
Solve Quadratic Equations which Contain 
Two Unknowns 254 



FUNDAMENTALS OF 
HIGH SCHOOL MATHEMATICS 

CHAPTER I 

A NEW WAY TO REPRESENT NUMBERS 

Section 1. It saves time to use abbreviations and let- 
ters, instead of words, to represent numbers. In order to 
save time in reading and writing numbers in your studies 
in arithmetic, you have already found it convenient to use 
certain abbreviations or letters to represent numbers. For 
example, instead of " dozen" you have used " doz." to stand 
for 12 ; C to stand for 100 ; M for 1000 ; cwt. (hundred- 
weight) for 100 lb., mo. (month) for 30 days, etc. It is 
necessary that we learn more about this new way of repre- 
senting numbers by letters because we shall use it in all our 
later work in mathematics. 

EXERCISE 1 

PRACTICE IN USING ABBREVIATIONS AND LETTERS TO REPRESENT 

NUMBERS 

1. How many eggs are 6 doz. eggs and 2 doz. eggs ? 

2. How many days in 3 mo. and 2J mo. ? 

3. Change 5 yr. and 2 mo. to mo. 

4. If 1 ft. = 12 in., change 4 ft. + 5 in. to in. 

5. If R (ream) stands for 500 sheets of paper, how 
many sheets in 2 R + 3 R ? 

6. Change 5 yd. — 2 ft. — 3 in. to in. 

7. Using d for 12, how many eggs in 2 d + 3 d 
eggs ? 



2 Fundamentals of High School Mathematics 

8. How many sheets of paper in 5R + 3 R — 6 R 
sheets ? 

9. Change 5;;/ + 4w-6w + m to smaller time 
units, that is, to " days/' if m = 30 days. 

10. Change 1 y (yards) + 6/ (feet) to smaller units, 
that is, to " inches." 

11. How many cents in 4 q + 6 d> if q and d stand 
for the number of cents in a quarter and dime 
respectively ? 

12. If h (hour) equals 60 m (minutes), and m equals 
60 s, (seconds), how many s in 2 k +3m? 

In these examples, you have used abbreviations or single 
letters to represent numbers or known quantities. We 
make use of abbreviations or single letters instead of 
•- because You 

State the reason here 

need a great deal of practice in doing this. The next ex- 
ercise will give more practice in representing numbers by 
letters in different kinds of examples. 



EXERCISE 2 
FURTHER PRACTICE IN USING LETTERS FOR NUMBERS 

1. Change 10 y + 4/ to inches (smaller units) if 
y and / stand for the number of inches in a 
yard and in a foot, respectively. 

2. Express 2/i + 5 m in seconds. 

3. What will 4 R + 3 R + R sheets of paper cost 
at \$ per sheet? 



A New Way to Represent Numbers 



4. 
5. 



At 4^ each, what will 1 d —2d eggs cost? 



10. 



11. 



12 



3f 



The length of the rec- 
tangle in Fig. 1 is 
represented by the 

expression 3f, and the 2 f 

width by the expres- 
sion 2/. What ex- 
pression will represent 
the perimeter? How 
many inches in the perimeter Uf = 12 inches? 

6. Change 1 p + 5p — 8p + Sp to ounces, if p 
stands for the number of ounces in one pound. 

7. The expression 14 y + 8 m + 5 d represents the 
age of a pupil in an algebra class. Express this 
pupil's age in days or as a certain number of d. 

8. Find the cost of 4^ T of coal at 30 cents per 
cwt., using the relation, T = 20 cwt. 

9. If d = 4 q, find how many q in 3 d -f- 5 d. 



li y = 12 m, 
2y + 4 /^ ? 



and in = 30 d, how many ^/ in 



A boy earned 27 dollars in a month ; his father 
earned n dollars. How many dollars would 
both earn in 3 months, if n stands for 60 dollars? 



If r= 3 t and /= 6 s, how many s in 4 r — 5/? 

13. Write the relation between y and m (*".*. year 
and month); between / and i (foot and inch); 
between T and cwt (ton and hundredweight); 
between h } m y and s {i.e. hour, minute, and 
second). 



4 Fundamentals of High School Mathematics 

In this you always express " how many" of one unit 
equals a "certain number" of the other unit. For example, 
1/= 12 inches, or, more abbreviated, 

Are yards, feet, and incites related units ? Are 
hours, minutes, and seeonds related units ? Are 
pounds and dollars related units ? Are dollars 
and cents related units ? 

Section 2. Word statements about quantities may be 
much more briefly expressed by using a single letter to 
represent a quantity. In the last section we saw that it 
saved time to use abbreviations or letters to represent 
quantities. Now we shall show that entire word state- 
ments about quantities may be expressed much more 
briefly by using a single letter to represent a number. 
To illustrate, consider next the four different ways of 
writing the statement of the same example. 



Illustrative example. 



(a) The "word" method of 
stating the example. 

(b) An abbreviated way to 
write it. 

(c) A more abbreviated way 
to write it. 

(d) The best way to write it. 



(a) There is a certain number 
such that if you add 5 to 
it the result will be 18. 
What is the number ? 

(&) What no. plus 5 equals 18? 

(c) No. + 5 = 18. 

(d) n + 5 = 18. 



It is clear that in all these cases the number is 13, and 
that it is most easily represented by the single letter n. 
Thus, the fourth method, n + 5 = 18, illustrates the very 



A New Way to Represent Numbers 5 

great saving that is obtained by the use of single letters 
for numbers. This method will be used throughout all 
our later work. This is one of the aims of mathematics : 
to help you solve problems by better methods than you 
knew in arithmetic. 

EXERCISE 3 

Express the following word statements in the briefest 
possible way, using a single letter to represent the quantity 
you are trying to find. 

Illustrative example. If a certain unknown number be increased 
by 7, the result will be 16. This may be most easily written : 

n + 7 = 16, 
n = d. 



1. There is a certain number such that if you add 
12 to it, the result will be 27. What is the 
number ? 

2. If John had 7 more marbles, he would have 18. 
How many has he ? 

3. If the length of a rectangle were 5 inches less, 
•it would be 21 inches long. What is its length ? 

4. A certain number increased by 12 gives as a 
sum 35. What is the number ? 

5. The sum of a certain number and 7 is 18. Find 
the number. 

6. Three times a certain number is 21. Find the 
number. 

Explanation : Again, to save time, we agree that 
3 times n (8 times p, or 12 times x, etc.) shall be 



Fundamentals of High School Mathematics 

written 3 • n or, more briefly, 3 n. Understand, 
therefore, that whenever you meet expressions 
like 8/>, 12 x, 17 y, etc., they mean multiplication, 
even though no " times'' sign ( x ) is printed. 
In the same way ~ of a certain number is written 

nor n ; of a certain number, - or - n. 
3 3 2 2 2 

7. Two thirds of a certain number is 10. Find the 

number. 

8. Three fourths of a certain number is 15. Find 
the number. 

9. The difference between a certain number and 
5 is 9. What is the number ? 

10. The difference between a certain number and 
12 is 13. What is the number? 

11. The sum of 16 and a certain number is 29. 
Find the number. 

12. Three fifths of a certain number is 27. What 
is the number ? 

13. The product of 11 and a certain number is 77. 
Find the number. 

14. If your teacher had $50 less, he would have 
I 15. How much has he ? 

15. The quotient of a number and 7 is 3. What is 
the number ? 

16. If the area of a rectangle were increased 
12 sq. ft., it would contain 40 sq. ft. What is 
its area ? 



A New Way to Represent Numbers 7 

17. 13 exceeds a certain number by 4. What is 
the number ? 

Explanation : Does this mean that 13 is larger, or 
smaller, than the certain number ? How do you 
determine how much larger one number is than 
another ? 

18. Two thirds of the number of pupils in a class 
is 28. How large is the class ? 

19. Tom lacked $7 of having enough to buy a $50 
Liberty Bond. How much did he have ? 

20. Three times a certain number plus twice the 
same number is 90. Find the number. 

21. The difference between 20 and a certain num- 
ber is 4. What is the number ? 

Explanation: It seems most consistent to inter- 
pret, " the difference between two numbers," as mean- 
ing, " the first number minus the second number." 1 

22. The number of pennies Harry has exceeds 30 
by 7. How many has he ? 

23. The sum of two numbers is 40. One of them 
is 27. What is the other ? 

24. The difference between two numbers is 21. 
The larger is 60. What is the smaller ? 

25. The product of two numbers is 95. One is 5. 
What is the other ? 

26. The quotient of two numbers is 13. The di- 
visor is 5. Find the dividend. 

In these exercises you have been changing, or translat- 
ing, from the language of ordinary words into algebraic 
language ; you have been making algebraic statements out 



8 Fundamentals of High School Mathematics 

of word statements. The essential thing in this transla- 
tion is the representation of numbers by letters. We should 
note carefully also that we have begun the practice of 
using a letter to stand for a number which is unknown. 
( )i course in these simple examples there is only one un- 
known number and it is easy to see at a glance each time 
what it is. 



HOW TO REPRESENT TWO OR MORE UNKNOWN NUM- 
BERS, WHEN THEY HAVE A DEFINITELY KNOW 
RELATION TO EACH OTHER 

Section 3. In the examples which you have worked in 
preceding lessons, you have had to represent only one 
number in each problem. To illustrate, in Example 9, 
Exercise 3, as in all the other examples solved thus far — 
"the difference between a certain number and 5 is 9." 

Only one number has to be represented. But in most 
of the examples that you will meet in mathematics you 
will have to represent two or more numbers which have a 
definitely known relation to each other. For example, 
consider this problem : 

Suppose Tom has 5 times as many marbles as John 
has. How many do they both have ? 

It is clear that there are two numbers to be represented; 
namely, the number that Tom has and the number that 
John has. Furthermore, since there is a definite relation 
between these two numbers, that is, one is 5 times the 
other, it is important to see that each can be represented 
by the use of the same letter. 

If you let n stand for the number John has, what must 
represent the number Tom has ? Since the example states 



* 



A New Way to Represent Numbers g 

that Tom has 5 times as many as John, then Tom must 
have 5 n marbles. In the same way, together they have 
the sum of the two ; namely, n + 5 n, or 6 n. 

The best way to state this, however, in algebraic lan- 
guage, is to use a set form like the following : 

Let n = the number John has. 
Then 5 n = the number Tom has, 
and 5 n -f n, or 6 n = the number both have. 

The next exercises will show how two or more unknown 
numbers may be represented by using the same letter, if 
the numbers have a definite relation to each other. 



EXERCISE 4 

1. Harry has four times as many dollars as James 
has. If you let n stand for the number of dol- 
lars James has, what will stand for the number 
Harry has ? for the number they together 
have ? 

2. The number of inches in a rectangle is 7 times 
the number in its width. If n stands for the 
number oi inches in its width, what will repre- 
sent the number in its length ? in its perim- 
eter ? 

3. An agent sold three times as many books on 
Wednesday as he sold on Tuesday. Represent 
the number sold each day. State algebraically 
that he sold 28 books during both days. 

4. There are twice as many boys as girls in a 
certain algebra class. If there are ;/ girls, how 
many boys are there? How many pupils? 



10 Fundamentals of High School Mathematics 

State algebraically that there were 36 pupils in 
the class. Find the number of boys. 

5. Oil a certain day Fred sold half as many papers 
as his older brother. How can you represent 
the number each sold ? the number both sold ? 

6. During a certain vacation period there were 
three times as many cloudy days as clear days. 
Express the number of each kind of days, and 
the total number of days. If the vacation con- 
sisted of 60 days, how many days of each kind 
were there ? 

7. A rectangle is three times as long as it is wide. 
If it is x feet wide, how long is it ? What is 
its perimeter ? 

8. If one side of a square is s inches long, what is 
the perimeter of the square ? State algebraically 
that the perimeter is 108 inches. 

9. The sum of three numbers is 60. The first is 
three times the third, and the second is twice 
the third. If n represents the third number, 
what will represent the first ? the second ? their 
sum ? State algebraically that the sum is 60, 
and then find each number. Why do you 
think it was advisable to represent the third 
number by n ? 

10. John sold five times as many papers as Eugene. 
If n represents the number Eugene sold, what 
will represent the number John sold ? What 
expression will represent the difference in the 
number sold? Make a statement showing that 
this expression is 80. 



A New Way to Represent X timbers n 

11. A farmer sold four times as many dollars' 
worth of wheat as of corn. If he received x 
dollars for the corn, what will represent the 
amount he received for both ? 

12. A has 11 dollars. B has three times as many as 
A, and C has as many as both A and B. What 
will represent the number of dollars all three 
together have ? 

13. A horse, carriage, and harness cost 6500. The 
carriage cost three times as much as the har- 
ness, and the horse twice as much as the car- 
riage. If you let n represent the number of 
dollars the harness cost, what will represent 
the cost of the carriage ? of the horse ? of all 
together? Make an algebraic statement show- 
ing that all three cost 6500. Can you now find 
the cost of each ? 

14. A man had 400 acres of corn and wheat, there 
being 7 times as much corn as wheat. Show 
how the number of acres of each could be rep- 
resented by some letter. Make an algebraic 
statement showing that he had 400 acres of 
both. 

15. The rectangle shown in 
Fig. 2 is three times as 



long as wide. State al- FlG 2 

gebraically that the per- 
imeter is 64 in. What are its dimensions ? 
In the problems just studied you have been considering 
two or more numbers which had a definite RELATION to 
each other and each of which had to be represented by 
using the same letter. For example, you had to note that 



12 Fundamentals of High School Mathematics 

one number was always a certain number of times another 
one, or was a certain part of another one. In each prob- 
lem you had to decide which of the unknown numbers you 
would represent by that letter. In general it is best to 
represent the smaller of the unknown numbers by n or 
by p or by any letter. The other numbers must then be 
represented by using the same letter which you selected to 
represent the first one. 

EXERCISE 5 

Write out the solution of each of the following. Be 
sure to use the complete form illustrated below, in solving 
each example. 

Illustrative example. The larger of two numbers is 7 times the 
smaller. Find each if their sum is 32. 

Let s = smaller no. 
Then 7 s = larger no. 
and 7 s + s = 32, 
or 8 s = 32, 
or 5 = 4, 
and 7 5 = 28. 



1. William and Mary tended a garden, from which 
they cleared $72. What did each receive if it 
was agreed that William should get three times 
as much as Mary? 

2. The perimeter of a rectangle is 48 inches. Find 
the dimensions if the length is 5 times the 
width. 

3. The sum of three numbers is 60. The second 
is twice the first, and the third equals the sum 
of the first and second. Find each. 

4. Divide $48 between two boys so that one shall 
get three times as much as the other. 



A New Way to Represent X umbers 13 

5. Twice a certain number exceeds 19 by 5. Find 
the number. 

6. The product of a certain number and 5 is 35. 
Find the number. 

7. A man is twice as old as his son. The sum of 
their ages is 90 years. Find the age of each. 

8. The sum of three numbers is 120. The second 
is twice the first and the third is three times 
the first. Find each. 

9. The perimeter of a certain square is 14-4 inches. 
Find the length of each side. 

10. The perimeter of a rectangle is 160 inches. It 
is three times as long as it is wide. Find its 
dimensions. 

11. William is three times as old as his brother. The 
sum of their ages is 36 years. How old is each ? 

12. One number is five times another. Their dif- 
ference is 10. Find each. 

13. The sum of three numbers is 14. The second 
is twice the first, and the third is twice the sec- 
ond. Find each number. 

14. One number is eight times another. Their dif- 
ference is 60. Find each. 

15. A rectangle (Fig. 3) ~ ^ 
which is formed by 
placing two equal 
squares together has 
a perimeter of 150 
feet. Find the side ~ 7C "~~xT 
of each square, and Fig. 3 
the area of the rectangle. 



X 



14 Fundamentals of High School Mathematics 

16. Three men, A, B, and C, own 960 acres of land. 
B owns three times as many acres as A, and 
C owns half as many as A and B together. 
How many acres has each ? 

17. John sold half as many thrift stamps as Harry 
sold ; Tom sold as many as both the other boys 
together. Find how many each sold, if all sold 
144 thrift stamps. 

18. Divide #21 among three boys, so that the first 
boy gets twice as much as the second, and 
the second boy gets twice as much as the 
third boy. 

Section 4. The most important thing in mathematics : 
the EQUATION. In all the examples in Exercise 5 you 
have translated word sentences into algebraic sentences. 
These algebraic sentences are always called equations. 

They are called equations because they show that one 
number expression is equal to another number expression. 
For example, you have stated that n + 5 = 18. This 
statement merely expresses equality between the number 
expression n + 5, on the left side of the = sign, and the 
number expression 18, on the right side. 

Furthermore, you have been finding the value of the un- 
known number in each of these equations. From now 
on, instead of saying "find the value of the unknoivn in an 
eqtiaiion" we shall say : " solve the equation." For 
example, if you solve the equation 

7s + s = S2, 

you " find the value " of s ; namely, s = 4. 



A New Way to Represent Numbers 15 

EXERCISE 6 

Solve the following Equations : 

1. ^> + 3=8. This might be written: What no. 

+ 3=8, or ? + 3 = 8. 

2. x — 5 = 10. This might be written : What no. 

- 5 = 10, or ? - 5 = 10. 

3. 2 n = 25. This might be written : 2 times ? = 25. 

4. 5 a = 275. This might be written : 5 times 
? = 275. 

5. l*- = 7. This might be written: \ times 
?=7. 

6. § c = 12. This might be written : | times 
? = 12. 

It is always helpful to think of an equation as 
asking a question. Thus, 5 a + 1 = 16 should be 
thought of as the question : 5 times what number 
plus one gives 16 ? 

7. 2£ + l = 21 14. 3*+ 1=16 

8. 5^-3 = 27 15. 7£-2 = 12 

9. 4* = 13 16. 12^ = 27 

10. 12 = 3/ 17. 16 = 5j/-l 

11. 5 + * = 11 18. 6/ + 3/ = 27 

12. 6 — « = 2 19. 13 = 5 y 

13. 2/ + 3/ = 35 20. 21 = 5.r+l 

21. 4* =17 

Section 5. Translation from algebraic expressions into 
word expressions. In the previous work you have trans- 
lated from word statements into algebraic expressions. It 
is also very helpful to translate the algebraic expressions 
back into word expressions. For example, n + 1 = 13 is the 



16 Fundamentals of High School Mathematics 

same as the word statement " the sum of a certain number 
and 4 is 13." In the same way, the algebraic statement 
4 y = 2(> should be translated as follows : 

11 the product of a certain number and 4 is 20," or 
"four times a certain number equals 2b\" 

The next exercise will give practice in this important 
process, i.e. translating from algebraic statements into word 
statements. 

EXERCISE 7 

Translate each of the following algebraic statements 
into word statements : 



1. 

2. 


y + 4 = 20 

2/; + l = 31 


7. 


| + 1 = 8 

o 


12. 

13. 


^-4 = 12 

a-b = l 


3. 

4. 


13 = 2+jy 
2tf + 3# = 55 


8. 
9. 


5x = 18 


14. 


d 


5. 


I n + 1 = 12 


10. 


bh = 20 


15. 


\*i-\n = 


6. 


;/ + 3// = 24 


11. 


h = 60 m 







SUMMARY OF CHAPTER I 

After studying this chapter you should have clearly in 
mind : 

1. It saves time to represent numbers by letters. 

2. Worded problems may be translated into algebraic 
statements. 

3. Equations are statements that two numbers or two 
algebraic expressions are equal. 

4. Solving equations means finding the value of the 
unknown number or letter in the equation. 

5. Algebraic expressions may be translated into word 
expressions. 




CHAPTER II 

HOW TO CONSTRUCT AND EVALUATE FORMULAS 

Section 6. Further need for abbreviated language : Short- 
hand rules of computation. In this chapter we shall study 
abbreviated or shorthand rules for solving problems. 
People who have found it necessary to compute over and 
over again the areas or perimeters of such figures as rec- 
tangles, triangles, circles, etc., have found it very conven- 
ient to abbreviate the rules for solving these problems into 
a kind of shorthand expression which can be more easily writ- 
ten or spoken than the long rules. 
For example, suppose you wanted 
to make a complete statement, 
either in writing or orally, concern- 
ing how to find the area of the 
rectangle which is represented by 
Fig. 4. You might express it, as 
you did in arithmetic, as follows : 

(1) The number of square units in the area of a rectangle is the 
number of units in its base times the number of units in its 
height. 

This long word rule can be greatly shortened by using 
abbreviations or suggestive letters to represent the number 
of units in each of its dimensions. Thus, a shorter way of 
expressing this rule is : 

(2) Area = base x height. 

A third and still more abbreviated way of expressing it is : 

(3) A = b x //, 

in which A, b, and // mean, respectively, the number of 
units in the area, base, and height. And finally, remem- 
bering that b X h is usually written as bh y the entire state- 
ment becomes : 

(i) A = b/i. 
17 



18 Fundamentals of High School Mathematics 

This last statement tells us everything that the first 
statement did, and requires much less time to read or to 
write. Such algebraic expressions are called formulas. 

Section 7. What is a formula? From the previous illus- 
tration we see that a formula is a shorthand, abbreviated 
rule for computing. We must remember, however, that 
the formula A = bJi is, at the same time, an equation. 
Since it is an equation that '^frequently used, and which 
always appears in that particular form, we have come to 
call it a FORMULA. 



I. COMPUTATION OF AREAS AND PERIMETERS BY 
FORMULA 



EXERCISE 8 
COMPUTATION OF THE AREA OF RECTANGLES BY THE FORMULA 

l. Illustrative example. Find 
the area of a rectangle (Fig. 5) in 
which b= 10 and h = 7.5, using 
the formula 

A = bh. 

Solution : (1) A = bh. 

(2) A= 10 x 7.5. 

(3) A = 75. 




2. Find A when b = 8.25 and h = 4. 

3. Find A when h = 4.5 and b = 12. 

4. Find b when A = 50 and h = 5. 

5. What is A if b = 6.5 and h = 5.4 ? 

6. What is h if A = 40 and b = 6| ? 

7. Find A if h = 2.5 and b = 6.4. 

8. What is b if A = 450 and // = 22.5 ? 



How to Construct and Evaluate Formulas 19 

9. If A = 200 and b = 7.5, what does h equal ? 

10. If A = 625 and h = 50, what does b equal ? 

11. What is ^4 if b = 40 and h is twice as large as b ? 

12. Find ^ if h = 16.2 and * = \ of A. 

13. b == 12 and ^ = f £. What is ^ ? 

14. Find ^ if h = 20 and A + b = 32. 

Section 8. Perimeters of rectangles. In Section 7 we 
saw that it was convenient to use a formula for the area 
of a rectangle. In the same way 
it is helpful to have a formula for 
the perimeter of any rectangle. 

Since' the perimeter of any 

rectangle is the sum of the bases 

and altitudes, the shortest way to 

... J Fig. 6 

express this is : 

(1) Perimeter = 2 x base plus 2 x height, 
or by the formula 

(2) P = 2 b + 2 h. 

EXERCISE 9 

COMPUTATION OF THE PERIMETERS OF RECTANGLES BY THE FORMULA 

1. Illustrative example. Find the perimeter if the base 
is 13 and the height is 9 ; or, more briefly, find P if b = 13 
and h = 9. 

Solution: (1) P=2b + 2h. 

(2) P=2-13 +29. 

(3) P = 26 + 18 = 44. 




2. Find P if h = 10.5 and * = 9. 

3. Find P if >fc = 18.4 and b = 12.8. 

4. What is kiiP = 40 and b = 10 ? 



2o Fundamentals of High School Mathematics 



What is b if P = 60 and h = 14? 
Find A if />= 18.4 and * = 4.6. 
If P = 110 and A = 22.5, what is b? 
What is P if h = 18 and * = 2 A? 
/> = 1 00. Find * and h if * = //. 



What is h\lP = 120 and * = £ P ? 



5. 
6. 
7. 
8. 
9. 
10. 

Section 9. The formula for the area of any triangle. 

What is the area of this triangle if its base is 12 ft. and its 
height is 8 ft. ? How do you find 
the area of any triangle if you 
know its base and altitude ? 

Show that the most economi- 
cal way to state this rule, or re- 
lation, between the area, the 
base, and the height, is by the 

formula A = — . FlG# 7 

The examples in the following exercise will give you prac- 
tice in using this important formula. 




EXERCISE 10 
COMPUTATION OF THE AREA OF TRIANGLES BY THE FORMULA 

l. Illustrative example. 

Find the value of A if b = 22 and h = 12. 



Solution: (1) A = — . 
v 2 



(2)A = 



22 x 12 264 



= 132. 



Write your work in a neat, systematic form. 

2. Find the value of A if b = 18 and h = 6|. 

3. What is the value of A if b = 12.5 and // = 20? 



How to Construct and Evaluate Formulas 21 

4. If h = 16.8 and b = 28, what does A equal ? 

5. What is the value of b if A = 300 and h = 50 ? 

6. Find h if A = 240 and * = 20. 

7. Determine £ if ^4 = 100 and h = 15. 

8. What is AHA = 6.25 and * = 10.5 ? 

9. Find the value of A if b = 22 and /* = T 6 T £ 

10. What is b if ^ = 120 and h = \ A ? 

11. /z = 20 and b — \k. What does ^ equal? 

12. Can you find b and // if ^4 = 256 and b = 2/i? 

13. Two triangles have equal bases, 10 in. each, but 
the height or altitude of one is twice that of the 
other. Are their areas equal? Show this by 
an illustration. 

14. What change occurs to A if b is fixed in value, 
but if h gets larger ? What is the relation be- 
tween A and h if b is fixed ? 

Section 10. The formula for the circumference of any 
circle. You will recall from arithmetic the following state- 
ment for the circumference of a 
circle : 

" The circumference of a circle 
is obtained by multiplying 
twice the radius by 3.1416." 
With our new method of using letters 
instead of words or numbers, this is 
much more briefly expressed by the 
formula FlG - 8 

C = 2 irR. 

7r is a symbol used to represent the number 3.1416. Use 
this value for it in the problems which follow. 




22 Fundamentals of High ScJiool Mathematics 

I 

EXERCISE 11 
COMPUTATION OF THE CIRCUMFERENCE OF CIRCLES BY THE FORMULA 

1. What is the value of C if R = 12 ? 

2. What does C equal if R = 5| ? 

3. Find the value, of R if C= 31.416. 

4. Determine R when C= 100. 

5. Find C if the diameter of the circle is 16.4. 

6. What is the value of C if R = ^ ? 

7. A Boy Scout wishes to make a circular hoop 
from a piece of wire 14 ft. long. Determine 
the radius of the largest possible hoop he can 
make. 

8. The radius of one circle is 5, and the radius of 
another is twice as great. Find the circumfer- 
ence of each, and note whether one circumfer- 
ence is twice the other. 

9. Think of a circle of some particular radius. 
Then imagine that the radius begins to increase. 
What happens to the circumference ? Is there 
any particular connection or relation between C 
and R ? 

II. "EVALUATION": HOW TO FIND THE NUMERICAL 
VALUE OF AN ALGEBRAIC EXPRESSION 

Section 11. In the examples which you have just solved 
wc have used the long expression " What is the value of" 
or " Find the value of " in referring to the particular letter 
which was to be found. Instead of these long expressions 
we shall now use the single word evaluate. It means 
exactly the same thing as the longer expression. Thus 



How to Construct and Evaluate Formulas 23 

to evaluate an algebraic expression means to find its nu- 
merical value, exactly as in the previous examples. This 
is done by " putting in" or by substituting numerical values 
for the letters. A few examples will make this clear. 



EXERCISE 12 
EVALUATION OF COMMONLY USED FORMULAS 

1. Evaluate A = ^ if b = 10 and h = 14.6. 

2. Evaluate, or find the value of, P in the expres- 
sion P = 2 6 + 2 h if £ = 26 and h = 12.4. 

3. Evaluate C = 2 irR if R = 14. 

4. Evaluate F = /ze>// if /= 10, w = 6J, and // = 5. 

5. Find the value of i in the formula i=prt if 
/ = $640, r = yo"o> an d * = 4. 

6. Evaluate A = ttT? 2 if R = 6. 

7. Evaluate ^ = - if E = 110 and i? = 10.5. 

R 

8. What is h in the algebraic expression P = 2d 
+ 2kifP=8Q and £ = 12.8? 



III. THE USE OF EXPONENTS TO. INDICATE 
MULTIPLICATION 

Section 12. Need of short ways to indicate multiplica- 
tion. A very large part of our work in mathematics is 
that of finding numerical values. In many of our problems, 
therefore, we shall need short ways of indicating multipli- 
cation. For example, in arithmetic, the multiplication of 
5 x 5 is sometimes written as 5 2 ; or the multiplication of 



24 Fundamentals of High School Mathematics 



6 x 6 x 6 as 6 8 . In algebra, to save time, this notation, or 
method of indicating multiplication, is always used. Thus, 
instead of writing" b X b or u x u X u we will write b 2 or /z 3 . 
This little number that is placed to the right of and above 
another number tells how many times that number is to 
be used as a factor. These numbers are called exponents. 
Numbers with exponents are read as follows : 

3 a 2 means 3 times a times a, and is read "3a square.' 11 
This does NOT mean 3 a times 3 a. The exponent affects only 
the a. 

5 #* means 5 times b times b times b, and is read " 5 b cube." 
This does NOT mean 5 b times 5 b times 5 b. The exponent 
affects only the b. 

Here, as well as throughout all later mathematical work, 
you will need to be able to evaluate algebraic expressions 
which involve exponents. For 
example, the area of the rec- 
tangle shown here is the expres- 
sion 3 IV 2 , which is obtained by 
multiplying 3 W by W. Now 
the numerical value of this area 
depends upon the value of W ; that is, if W is 4, then the 
area is 3-4-4, or 48 ; but if W is 2, then the area is 
3-2-2, or 12. In the same way 
the volume of the rectangular box in 
Fig. 10 is represented by the ex- 
pression 2x s , or 1x • x • x. Again, 
you see that the numerical value of 
the volume depends upon the value 
of x. Thus, if x is 5, the volume is 
obtained by evaluating the expression 2x s , which gives 
2 • 5 • 5 • 5, or 250. The next exercise gives practice in 
evaluating algebraic expressions containing exponents. 






-3W- 



T 



Fig. 9 




Fig. 10 



How to Construct and Evaluate Formulas 25 

EXERCISE 13 
PRACTICE IN EVALUATION 

1. Illustrative example. 

Evaluate 2 a& 2 + 3 a 2 b + ac, if a — 4, b = 3, and c = 1. 
Solution : 2- 4- 3- 3 + 3-4. 4- 3 + 4-1 = 72 + 144 + 4 

= 220. 
Note that the numbers are substituted for, or put in place of, 
the letters. 





a + b — c 


9. 


aHc* 


10. 


M+ 1 

a o c 


11. 


a s — b s — c z 


2 a 2 


12. 


a h + b a + c a 



Using the values of a, b, and c given in Example 1, 
evaluate each of the following expressions : 

2. a 2 +d 2 + c 2 a-b + c 

3. 3 abc 

4. a 2 b + ab 2 

5. ac 2 + cb 2 + ba 2 

6. a 3 + b z + c 3 

7. ? + b - + C - 
b a b 

13. The formula d= 16 / 2 tells how far an object 
will fall in any number of seconds. Find how 
far a body will fall in 1 sec. of time, that is, 
when t = 1, Do you believe it ? How could 
you test it ? 

14. Using the formula in Example 13, find how far 
an object will fall in 2 sec. of time, that is, when 
t = 2. How could you test the truth of this? 

15. The horsepower of an automobile is given by 

ZM 72 
the following formula: -^~~y in which D rep- 

2.5 

resents the diameter of the piston, and N the 



26 Fundamentals of High School Mathematics 

number of cylinders. What is the horsepower 
of a Ford, which has 4 cylinders, and in which 
D = 31 in. ? 



IV. THE CONSTRUCTION OF FORMULAS 

Section 13. It is very important to be able to make a 
formula for any computation that must be performed over 
and over again. For example, we often have to find the 
area of a square. Instead of saying or writing each time 
" the area of a square is equal to the square of the number 
of units in one of its sides," it saves time to use the for- 
mula A = s 2 , in which A = area and s = one of the 
sides. This formula tells all that the word ride says arid 
requires much less effort. To give practice in this kind of 
work, construct a formula for each of the examples in the 
following exercise. 

EXERCISE 14 

1. (a) Find the volume of a rectangular box whose 

dimensions are 12, 8, and 6 inches. 
(b) Make a formula for the volume of any box. 

2. (a) What is the area of a circle whose radius is 

9 in. ? 
(b) Write the formula for the area of any circle. 

3. {a) How many square inches in the entire sur- 

face of a cube whose edge is 8 inches ? 
(b) Give a formula for the area of the entire sur- 
face of any cube. 
4! (a) Find the interest on %400 for 2 years at 6%. 
(b) Make a formula for the interest on any prin- 
cipal for any rate and for any time. 



How to Construct and Evaluate Formulas 27 

5. (a) How many cubic inches in a block 2 1 by 3' 

by 4'? 
(b) Make a formula for the number of cubic 
inches in any rectangular solid whose dimen- 
sions are expressed in feet. 

6. Make a formula for, or an equation which tells, 
the cost of any number of pounds of beans at 
12 cents per pound. 

7. What equation or formula will represent the 
area of any rectangle whose base is 5 inches, but 
whose height is unknown ? Evaluate your for- 
mula for h = 3.4. 

8. An automobilist travels 20 miles per hour. 
What formula or equation will represent the dis- 
tance he travels in / hours ? Evaluate this 
formula : t = 5 hr. 20 min. 

REVIEW EXERCISE 15 

1. What does an equation express ? Is 7 + 4 
= 6 + 6 an equation ? 

2. Does 2 n + 1 = 21, if n = 9, make an equation? 
if n = 10 ? 

3. The formula for the perimeter of a rectangle, 
p = 2 b +2b y contains three unknown numbers. 
How many of them must be known in order to 
use this formula to solve an example ? 

4. Read each of the following equations as ques- 
tions, and find the value of the unknown number: 
(a) ±y + 3 = 21 (d) 3 y - 5 = 16 

(6) 20 = 6 + 2x ( e ) x + x= 36 

(c) 5 c + 2 = 42 (/) &+ b + 1 = 23 



28 Fundamentals of High School Mathematics 

5. Three times a certain number, plus 2, equals 
38. Find the number. 

6. Donald saved twice as much money as his 
older brother. Express in algebraic language 
that both together saved $ 96. How much did 
each save? 

7. Evaluate the formula V=P(V=the volume 
of a cube), if /= 4.]. 

8. The first of three numbers is twice the second, 
and the third is twice the first. Find each 
number if their sum is 105. 

9. Construct a formula for the cost of any number 
of eggs at 30 cents per dozen. 

10. What is the difference in meaning between 10 n 
and ;/ + 10 ? Does 4 w mean the same as 4 + w ? 



SUMMARY 



The most important principles and methods which we 
have learned in this chapter are the following : 

1. A formula is merely a shorthand rule of 
computation. 

2. Formulas are " evaluated" or "solved" by sub- 
stituting numbers for the letters in the formula. 

3. Exponents are used as short methods of indicat- 
ing multiplication. An exponent of a number 
tells how many times that number is taken as a 
factor. 

4. We should construct a formula for any kind of 
problem which we have to solve frequently. 



CHAPTER III 

HOW TO USE THE EQUATION 

Section 14. The importance of the equation. Nothing 
else in mathematics is as important as the equation, and. the 
power to use it well. It is a tool which people use in stat- 
ing and solving problems in which an unknown quantity 
must be found. In the last chapter we saw that the 
formula^ or equation, was used to find unknown quantities, 
sometimes the area, sometimes the perimeter, etc. The 
fact that the equation is used as a means of solving such a 
large number of problems is the reason we shall study it 
very thoroughly in this chapter. 

Section 15. The equation expresses balance of numerical 
values. The equation is used in mathematics for the 
same purpose that the weighing " scale " is used by clerks ; 
that is, to help in finding some value which is unknown. 
The scale represents balance of weights : similarly the 
equation represents balance of numerical values. To un- 
derstand clearly the principles which are applied in dealing 




3<d Fundamentals of High School Mathematics 

with equations, we should consider the scale, as repre- 
sented in Fig. 11. In this case a bag of flour of un- 
known weight, together with a 5-pound weight, balances 
weights which total 25 pounds on the other side of the 
scale. 

Now, if ;/ represents the number of pounds of flour, it 
is clear that the equation 

7i + 5 = 25 

represents a balance of numerical values. Obviously, n is 
20, for the clerk wtmld take 5 pounds of weight from each 
side, and still keep a balance of weights. 

This principle, namely, that the same weight may be 
taken from each side without destroying the balance of 
weights, can be applied to the equation 

n + 5 = 25. 

That is, we may subtract 5 from each side of- the equation, 
giving another equation, 

^ = 20. 

This suggests an important principle that may be used in 
solving equations ; namely, — 

The same number may be subtracted from each side 
of the equation without destroying the equality, or 
balance, of values. 

If you take something from one side of the scale, or of 
the equation, what must you do to the other side ? Why ? 

The fact that the equation expresses the idea of balance 
makes it easy to reason about it, and find out all the things 
that can be done without changing the balance or equality. 
The next exercise suggests this kind of study of the equa- 
tion. 






How to Use the Equation 31 

EXERCISE 16 

By thinking of the equation as a balance, you should be 
able to complete the following statements. Fill in the 
blanks with the proper words. 

1. Any number may be subtracted from one side 

of an equation if 1 is ! from the 

other side. 

2. Any number may be added to one side of an 
equation if ! is I to the other. 

3. One side of the equation may be multiplied by 
any number, if the other side is I by the 



4. One side of an equation may be divided by any 
number, if the other side is I by the 



These are very important principles, and are used in 
solving any equation. They are generally called axioms. 
They must be understood and mastered. The examples, 
of the next exercise have been planned to help you learn 
how to apply them. 

EXERCISE 17 

In each of the following examples you can use one of 
the four principles stated above to explain what has 
been done, or to state the reason for doing it. Thus, if 
2,r=8, then x = i } because of the principle: 

" One side of an equation may be divided by a number 
if the other side is divided by the same number." 

For each example, you are to state the principle which 
permits or justifies the conclusion. 



32 Fundamentals of High School Mathematics 

1. If 4/; = :!:!, then what is done to each side to 
give b = 5 J ? 

2. If ]j' = 7, then to get y = 14, what do you do 
to each side ? 

3. If a - + 4 = 13, then to get x = 9, what do you do 
to each side ? 

4. If 5c = 32. 5, then what is done to each side to 
give c = 6.5 ? 

5. If 6a = 12, then what is done to each side to 
give 3 a = (> ? 

6. If y — 4 = 7, then what is done to each side to 
give y = 11 ? 

7. If x = 2 and j/ = 3, then why does x +y = 5 ? 

8. If b = 3 and £ = 10, then why does fo = 30 ? 

9. If x = 12 and jj/ = 4, then why does - = 3 ? 

y 

10. If a — 1 = 9, then why does # = 10 ? 

. ll. If b = 2/i, then what is done to each side to 
give 3 6 = 6 // ? 

12. If -ix + 3 = 23, then what is done to each side 
to give 4 x = 20 ? 

13. If oy — 3 = 27, then what is done to each side 
to give by = 30 ? 

14. If x + 7 = 19, then to make ;r = 12, what is done 
to each side ? 

15. If 2c — 4 = 8, then to make 2^ = 12, what is done 
to each side ? 

16. If 3 b + 1 = 22, then what is done to each side 

to give b = 7 ? 



How to Use the Equation 33 

17. If 5^ + 2 = 47, why does 5 b = 45 ? Then why 
does b — 9 ? 

18. If 6;r+2 = ;r + 22, then why does 5* + 2 = 22 ? 
and why does 5;r=20? and why does ;r = 4? 

19. If you know that ±w + 3 = w + 27, then why 
does 4 w = w + 24 ? 

and why does 3 w = 24 ? 

and why does w = 8 ? 
These examples are given to emphasize the fact that 
there are certain changes that can be made on both sides 
of an equation, without destroying the balance or equality. 
It should be clear that there must be some axiom or prin- 
ciple to justify every change that is made. 

EXERCISE 18 

Find the value of the unknown number in each of the 
following equations, telling exactly zvhat you do to eacli side 
of the equation. 



1. 


x + 5 = 13 


2. 


2, a = 11 


3. 


26 = 4j' 


4. 


2£ + l = 19 


5 « 


y -5 = 12 


6. 


J-* = 4.5 


7 - 


a r 

3 = 5 


8. 


2*- 3 = 17 


9. 


15 = x + 7 


10. 


2.r + 3;r = 35 



11. 


5y + -iy+y = 30 


12. 


15 = 3 

b 


13. 


5 c - 2 = 38 


14. 


27 = 6.r -3 


15. 


4 b + 7 = 47 


16. 


4j-j/ = 21 


17. 


5x + 1=23 


18. 


21* = 15 


19. 


f« = 18 


20. 


y 



34 Fundamentals of High School Mathematics 

21. 2£ + 3*=42 24. 18+* = 13 + 10 

22. *r-4 = 13 25. 7 + 2* = 23 + 10 

23. 2/;- 1=18 26. £*+£* = 18 

HOW TO CHECK THE ACCURACY OF THE SOLUTION OF 

AN EQUATION 

Section 16. When is an equation solved ? We have 
already noted that an cquatioii is solved when the numerical 
value of the unknown number is found. Thus, the equa- 
tion 4 a + 3 = 29 is solved when the numerical value of a 
is found. This leads to another very important question ; 
that is : How can you be certain your solution is correct ? 
In other words, how can you test or check the accuracy of 
your work ? 

For example, suppose that in solving the equation 
4 a + 3 = 29 
one member of your class obtains 8 for the value of a. Is 
his result correct ? There is only one way to be sure. 
That is to substitute or " put in " 8 in place of a in the 
equation, to see whether the numerical value of the left 
side equals the numerical value of the right side. In other 
words, does 

4 . 8 + 3 = 29 ? 

Clearly, not. Therefore, the solution is incorrect ; it does 
not check. Then what is the correct value of a ? Some 
of you doubtless think it is 6|. Let us test or check by 
substituting 6J for a, to see if the numerical value of one 
side of the equation will equal the numerical value of the 
other side. Does 

4.6,] +3 = 29? 

Yes. Then the equation is solved, or, to use the more gen- 
eral term, the equation is satisfied when a = 6|. 



How to Use the Equation 35 

Summing up, then, an equation is solved when a value 
of the unknown is found which satisfies the equation ; that 
is, one which makes the numerical value of one side equal 
to the numerical value of the other side. The solution of 
the equation is checked by substituting for the unknown 
number the value which we think it has. If, as the result 
of the substitution, we get a balance of values, then we 
know that the equation has been solved correctly. 

We have already had practice in substituting numerical 
values for letters. In Chapter II, we called this evalua- 
tion. Thus you see that each time you check the solution 
of an equation, you are evaluating the original equation, 

EXERCISE 19 
PRACTICE IN CHECKING THE SOLUTION OF EQUATIONS 

1. The pupils in a class tried to solve the equation 

6^-3 = 39. 
A few decided that a = 7, while the others in- 
sisted that a = 6. Which group was right ? Show 
how they could have checked or tested their re- 
sult. Why, do you think, some pupils got 6 for 
the value of a ? 

2. Does x = 5 in the equation 12 x — 7 = 10 x + 3 ? 
In other words, does x = 5 satisfy this equation ? 

3. Would you give full credit on an examination to 
a pupil who said that y = 4J would satisfy the 
equation 8y — 4 = 6 y + 3 ? Justify your answer. 

4. Show whether the equation b 2 + 5 b = 24 is satis- 
fied or solved if b = 3 ; if b = 2. 

5. Do you agree that the value of x is 6 in the equa- 
tion 10.*: — 4 = 58 ? Justify your answer. 



36 Fundamentals of High School Mathematics 

6. Is the equation - ^ + 6 = j b + 8 satisfied 

when /> = 8 ? 

7. Does -r = 24 satisfy the equation 

\x+\x + \x = t 2&} 

8. State in words how the solution of an equation 
is tested or checked. 

9. What is the value of learning to check very care- 
fully every kind of work you do ? 

EXERCISE 20 

Solve each of these equations. Write out your ivork for 
each one in the complete form illustrated in the first example. 
Check each one so that you can be absolutely certain that 
your work is correct. 

1. Illustrative example. 

6 b - 4 = 24. 

(1) By adding 4 to each side, we get 

6 b = 28. 

(2) By dividing each side by 6, we get 

b = 4|. 

(3) Checking, 

6 • 4f - 4 = 24. 
28 - 4 = 24. 



2. 


5 * - 2 = 38 


10. 


Sx + 2x+6x = 66 


3. 


6 £ + 3 = 45 


11. 


5 c + 3 = 78 


4. 


T x = x + 30 


12. 


1 v — 8 1 


5. 

6. 


22 = 5* + 2 


13. 


5 " 4 


7. 


2.]j/ + l =26 


14. 


10/; + 3 = 7£+15 


8. 


b + 5 b = 20 + * 


15. 


12*- 2 = 5*+ 26 


9. 


iy = lZ+y 


16. 


13 j/ = 2 j/ + 3 j/ +4^/ 



How to Use the Equation 37 

HOW TO GET RID OF FRACTIONS IN AN EQUATION 

Section 17. The use of the most convenient multiplier. 
In many equations that you have solved already it has been 
necessary to multiply each side of the equation by some 
number. For example, in \x — 10, it is necessary to mul- 
tiply each side by 2, which gives x = 20. Or, if you wanted 
to solve the equation 1^ = 3, it is necessary to multiply 
each side by 5, giving x = 15. 

But, suppose you had an equation 

would you get rid of both fractions by multiplying each side 
by 2 ? Would you get rid of both fractions by multiplying 
each side by 5 ? Here, as in all equations of this kind, 
you have to find some number which is a multiple of the 
different denominators. For this reason, in this example, 
10 is the most convenient number by which to multiply each 
term in the equation. 

Illustrative example. 

Multiplying each side of (1) by 10, we get 

10. i* -f 10.^x = 10- 14, 

or 5 x -f 2 x — 140, 

or 7 x — 140, 

or x = 20. 

Checking, by substituting the value of x in the original equation, 

gives 

1.20 + ^20=14, 
10 + 4 = 14. 



The study of this example shows that we can get rid of 
fractions in an equation if we multiply each side by the 
lowest common multiple of the denominators. We shall 
call this the most convenient multiplier. 



38 Fundamentals of High School Mathematics 



EXERCISE 21 

PRACTICE IN SOLVING FRACTIONAL EQUATIONS 

Solve and check each example. 

1. Illustrative example. * } n + \ n = 22. What is n ? 

(1) Multiplying each side by the most convenient multiplier, 12, 
gives 

12 • f n + 12 • J n = 12 • 22, 

or 8 n + 3 n = 264. 

(2) By adding 8 n and 3 n, we get 

11 n = 264. 

(3) By dividing each side by 11, we get 

72 = 24. 

(4) Checking, by substituting the value of n (24), in the original 
equation, 

22. 
16 + 6 = 22. 



2. What is the value of x in the equation 



\x+\x = m 



4. 



A man spent \ of his income for rent and \ for 
groceries. Using n to represent his income, 
make an equation which will state that he 
spent $ 660 for rent and groceries. Solve the 
equation. 

The dimensions of a rectangle are indicated 
on Fig. 12. What equation will state that the 
perimeter is 36 in. ? Solve the equation for /. 




Fig. 12 



How to Use the Equation 39 

5. 4 b + b = 1 j + 13. What is £ ? 

6. If three fourths of a single number be dimin- 
ished by one half of the number, the remainder 
is 10. Find the number. 

7. 2 X + % X -1 = 3. 

8. Three boys together had 65 cents. Tom had 
half as much as Harry, and Bill had two thirds 
as much as Harry. Translate this into an 
equation, and solve. 

9. ? + 2 + 5 =4 + ? What does n equal ? 

2 3 2 

10. One half of a certain number increased by 
four fifths of the same number gives 52 as a 
result. Find the number. 

11. Harry made two thirds as much money last 
year selling the Saturday Evening Post as John 
made ; Edward made three fourths as much as 
John. How much did each boy earn if all to- 
gether earned $ 145 ? 

12. ^- + |^-6 = 24. 

13. The sum of the third, fourth, and sixth parts of 
a number is 18. Find the number. 

Section 18. How word problems are solved by equations. 

It is important to note the principal steps involved in 
solving word problems. Let us take, as an illustration, 
Example No. 8 in Exercise 21. 

Three boys together had 65 cents. Tom had half as 
much as Harry, and Bill had two thirds as much as Harry. 
How much had each ? 



4-o Fundamentals of High School Mathematics 

1. The first important step in solving a word 
problem is to get in mind very clearly what is 
known and to recognize what is to be found 
out. In all problems some things are known 
and some things are to be determined. Thus, 
in this problem, we know how much money all 
the boys have together ; and we also know 
that Tom has half as much as Harry ; further- 
more, we know that Bill has two thirds as much 
as Harry. That is, we see that the statement 
of the amount that Tom and Bill each has 
depends upon the statement of the amount 
that Harry has. 

2. But we do not know how much Harry has. 
Then, as in all word problems, we represent by 
some letter, such as n, the number of dollars 
Harry has. In other words, the second step is 
to get clearly in mind what quantities are un- 
known, and to represent one of them by some 
letter. 

3. Next, all the parts or conditions of the prob- 
lem must be expressed by using the same 
letter. Thus, if Harry has n dollars, the 
number that Tom and Bill each has must be 
represented by using the same letter n, and 
not some other letter. That is, the word 
statement must be translated into an algebraic 
statement. It is always necessary, and usually 
difficult, to see that there must be a balance, an 
equality, between the parts of the problem. 
Thus, we must see that Harry's money, n t 
plus Tom's money, \ n y plus Bill's money, | n, 



How to Use the Equation 41 

must balance, or equal, 65 cents. This gives 
the complete algebraic statement : 

n + \ n^\- § n = 65. 

4. The equation which we have obtained must be 
solved. A value of the unknown must be found 
which will satisfy the equation. In this case n 
proves to be 30. 

5. Finally, the accuracy of the result must be tested 
by substituting the obtained value of n in the 
original word statement of the problem, to see if 
the statement holds true. 



EXERCISE 22 

Translate into algebraic language, and solve each of 
the following word statements. Check each one. 

1. Six more than twice a certain number is equal 
to 12. Find the number. 

2. Four times a certain number is equal to 35 di- 
minished by the number. What is the number? 

3. I am thinking of some number. If I treble it, 
and add 11, my result will be 32. What number 
have I in mind ? 

4. If fourteen times a certain number is dimin- 
ished by 2, the result will be 10. Find the 
number. 

5. What is the value of y in the equation 

4j/ + i?/ + 2 = 28? 

6. If seven times a certain number is decreased by 
8, the result is the same as if twice the number 
were increased by 32. Find the number, 



4.2 Fundamentals of High School Mathematics 

7. An algebra cost 12 cents more than a reader. 
Find the cost of each if both cost $1.64. 

8. The sum of the ages of a father and his son is 
57 years. What is the age of each if the father 
is 29 years older than the son ? 

9. The length of a school desk top exceeds its 
width by 10 inches; and the perimeter of the 
top is 84 inches. What are its dimensions? 

10. Divide $93 between A, B, and C, so that A 
gets twice as much as C, and B gets $10 more 
than C. 

11. A farmer sold a certain number of hogs at $20 
each, and twice as many sheep at $14 each. 
How many of each did he sell if he received 
$576 for all? 

12. Should a teacher give James full credit for the 
solution of the equation 

41^-7 = 3^ + 5 

if he obtained x=8^? Justify your answer. 

13. Make a drawing of a rectangle whose perimeter 
is represented by the expression 6 y + 20, writ- 
ing the dimensions on the drawing. 

14. The length of a rectangle is 5 inches more than 
twice its width ; its perimeter is 46 inches. 
What are its dimensions ? 

15. A school garden was 3| times as long as wide. 
To walk around it required 31 steps (27 in. 
each). Tell how to find its width, but do not 
actually find it. 



How to Use the Equation 43 

REVIEW EXERCISE 23 

1. The formula /i 2 = a 2 + b 2 is used in finding a side 
of a right triangle. Evaluate it if the base is 
13 and the altitude is 5. 

2. Make a formula for the number of revolutions 
made by the front wheel of a Ford car in going 
a mile, if the radius of the wheel is 14 inches. 

3. In what sense does the equation 7<5 — 5 = ^ + 25 
ask a question ? 

4. Give one illustration of the advantage of using 
letters for quantities. 

5. What are the four fundamental principles or 
axioms which are used in solving equations? 

6. Does x — \.\ satisfy the equation x 1 — 3 x = 6 ? 

7. Using m, s, and d for minuend, subtrahend, and 
difference, respectively, what equation or equa- 
tions can you make from them ? 

8. An autoist travels at an average rate of 24 mi. 
per hour. What distance will he cover in 2 hr. ? 
in 5 hr. ? in 10 hr. ? Make an equation or for- 
mula for the distance he will travel in t hr. 

9. Write a formula for the cost of any number of 
pounds of bacon at 30 cents per pound. 

10. Draw rectangles with bases of 2 inches each. 
What formula will represent the area of any 
such rectangle if // represents the height ? 

11. How do you get rid of fractions in an equation ? 
What is the most convenient multiplier in any 
particular equation ? 



44 Fundamentals of High School Mathematics 

12. When is an equation solved f 

13. If J/, ;;/, and/ stand for the multiplicand, mul- 
tiplier, and product, respectively, what formulas 
can you make from them ? 

14. If ;/ is a whole number, what is the whole num- 
ber next larger than n ? the whole number next 
smaller than ;/ ? 

15. In getting rid of fractions in the equation 

2 X ' 3 X == » 

show that 6 is a more convenient multiplier than 
12, 24, 18, 30, etc. 



SUMMARY 



From your study of this chapter, the following principles 
and methods should be kept clearly in mind: 

1. Equations express balance of value. 

2. If any change is made on one side of the equa- 
tion, the same change must be made on the other 
side. 

3. An equation is solved when a value of the un- 
known is found which satisfies the equation. 

4. The accuracy of your solution is checked by 
evaluating the equation for the value of the 
unknown. 

5. You can get rid of fractions in an equation by 
multiplying each side by the lowest common 
multiple of the denominators ; that is, by the 
most convenient multiplier. 



CHAPTER IV 

HOW TO REPRESENT THE RELATIONSHIP BETWEEN 
QUANTITIES WHICH CHANGE TOGETHER 

Section 19. The chief aim of mathematics. As we go 
about our daily work, we commonly deal with quantities 
which change together. For example, the cost of a rail- 
road ticket changes as the number of miles you travel 
changes ; that is, the cost and the distance change together. 
Or, the distance traveled by an autoist, if he goes at the 
rate of, say, 20 miles per hour, changes as the number of 
hours which he travels changes ; that is, the distance and 
the time change together. As a third illustration, suppose 
you wanted to make a trip of 100 miles. We know that 
the time required will change with or be determined by 
the way in which the rate changes ; that is, the time and 
the rate change together. 

The fact that we are always dealing with situations 
of this kind makes it necessary for us to know how to 
represent and determine these quantities which change to- 
gether, or which are related in some definite way. 
Mathematics shows us how to describe or express them. 
In fact, it is the chief aim of mathematics to help you to 
see how quantities are related to each other and to help 
you to determine their values. 

Section 20. The three methods of representing relation- 
ship. People have used three different methods for doing 
this: 

I. THE TABULAR METHOD 
II. THE GRAPHIC METHOD 
III. THE EQUATIONAL OR FORMULA METHOD 

This chapter will show how quantities, which are so re- 
lated W\2X they change together, can be represented by these 
methods. (// should be pointed out that it is not always 
possible to use the equational or formula method.) 

45 



46 Fundamentals of High School Mathematics 

To illustrate, suppose you wanted to tell some one about 
the temperature in Chicago on a certain July day. There 
are two ways to do this ; first, you might make a table, 
like the following, which would tell the temperature at 
each hour during the day. 

Table 1 

an illustration to show the way to tabulate temperature at 
different hours of the day 









A.M. 










P.M. 






Hour 


6 


7 


8 


9 


IO 


n 


12 


1 


2 


3 


4 


S 


6 


Temperature 


70 72 72 7& 80 83 85 90 96 


94 


86 76 77 



This method, which we shall call the tabular method, 
shows the way the temperature changes at different hours 
of the day. To understand the table, however, requires 
much more effort on the part of the reader than is re- 
quired to understand the second method, which is shown 
below. This pictorial or graphic method shows all that 
the tabular mctJiod shows, and has the advantage of being 
more easily interpreted. 



_. 1 r ..._ — — _ — 


100° " 
















< -^~-~*~ Sw 






y ' J**^ S - ... 


S :::::::::::::^-: :: :::::: : 5 — !::: 


Ill -r n o —"'^ 


^ 70 — -^*- 










60 -~ 



10 



12 
TIME 



Fig. 13. Graphic representation of temperatures at various hours 

of the day. 

Note that the horizontal line, or scale, shows the hours, 
or the time ; each large space represents one hour. The 



How to Represent Relationship between Quantities 47 

vertical line, or scale, shows the temperature ; on this scale 
each large space represents 10°, or each small space repre- 
sents 2°. Suppose we wanted to read from the graph 
what the temperature was at 11 o'clock. We find it by- 
looking along the time line, or time axis, until we come to 
the point marked 11 o'clock. We then look up, or down, 
to the line of the graph. In this case we have to go up 
to a point 11 \ small spaces above the 11 o'clock point. 
By looking back, to the left, to the vertical or temperature 
scale, we see that any point on this horizontal line stands 
for 83°. Hence the graph shows that at 11 o'clock the 
temperature was 83°. 

The following questions will help you compare the 
graphic and tabular methods of representing the relation 
between two numbers. 

EXERCISE 24 

In order to answer each of these questions, refer to the 
data of Table 1 and Fig. 13. 

1. Find, both from the table and from the graph, 
the highest temperature. 

2. What was the lowest temperature ? Which 
shows this the more easily, the table or the 
graph ? 

3. Between what hours did the temperature change 
the most rapidly ? 

4. About what do you think the temperature was 
at 9.30 a.m. ? 

5. Between what hours did the temperature change 
the least ? 

6. What might explain the rapid fall in temperature 
between 4 p.m. and 5 p.m. ? 



48 Fundamentals of High School Mathematics 

After answering the questions, are you not convinced 
that the graphic method gives the information which the 
reader may desire much more quickly and easily than the 
tabular method ? The fact that this is true has brought 
about a very wide use of graphic methods in all kinds 
of business and industry. Nearly every newspaper and 
magazine contains "graphs" of some kind. Your teacher 
will be glad to have you bring to class any graphs you may 
find in the newspapers or magazines. 

EXERCISE 25 

1. An east-bound train, running at 40 miles per 
hour, left Chicago at 8 a.m. Show from the 
graph, Fig. 11, how far the train was from 
Chicago at 10 a.m. ; at 11 a.m. ; at 11.30 a.m. ; 
at 2 p.m. At what time was the train 100 miles 
from Chicago ? 200 miles ? 











































250 


















































200 






























W 










U lrA 










£ 150 










*r* 










2 










H 










V) 










£ioo : 










Q 








































50 










m t 


?*~ 








J. 










s . 










r\ * 










<r 











8 



10 



11 12 

TIME 



Fig. 14. The line shows relationship between time spent and 
distance traveled. 



How to Represent Relationship between Quantities 49 

2. In Fig. 14 how many miles does each small 
space represent? How many hours does each 
large space equal ? 

3. In a newspaper the following graph, Fig. 15, 
was printed. It gives the prices of wheat, per 
bushel, from August 5 to August 10. What 
was the price on August 5 ? On August 7 ? On 
August 9 ? 



$1.40 


















































































































































































Wj.30 




















\, 














































u 










































































































































120 


















































































































































































110 






























i 



9 



10 



5 6 7 8 

AUGUST 

Fig. 15. Graphic representation of prices of 
wheat on various days of August, 1916. 

When was the price the highest ? the lowest ? 
Between what dates did the price change most ? 
change least ? 

Each large space on the vertical scale repre- 
sents how many cents ? What is measured 
along the horizontal scale ? What is the unit 
used on this scale ? 



SUMMARY OF IMPORTANT ASPECTS OF GRAPHIC 
REPRESENTATION 

Section 21. In the study of the previous examples, the 
following important aspects of graphic representation 
should be noted : 



50 Fundamentals of High School Mathematics 

1. Graphs always show the relation between two 
changing quantities; for example, they showed 
the relation between the number of miles 
traveled and the time required. 

2. Two rectangular axes are drawn. One of the 
changing quantities is measured on the hori- 
zontal axis ; the other changing quantity is 
measured on the vertical axis. 

3. These axes, or reference lines, are scales, 
marked off in a series of units. Thus, as in 
our illustrative examples, the horizontal axis 
may be a time scale, marked off into units of 
one hour each, and the vertical axis may be a 
distance scale, marked off into units of one 
mile, ox fifty miles, each. 

4. In making a graph one must choose units very 
carefully in order to be able to get all the 
information on the graph, and yet make it 
stand out as clearly as possible. 



EXERCISE 26 

1. The table below gives the earnings of a book 
agent for the latter part of July, 1915. Show 
the same thing graphically. 



Date 


19 


20 


21 


22 


23 


24 


25 


26 


17 


28 


Earnings -$ 


2.00 


3.50 


4.00 


5.00 


700 


4.50 


3.00 


8.00 


750 


5.00 



SUGGESTION. Represent time on the horizontal scale, and earn- 
ings on vertical scale. 



How to Represent Relationship between Quantities 51 



$2.40 



TIME 




Fig. 16. The line shows the relationship between the number 
of yards of cloth purchased and the total cost. 



2. Figure 16 is a price graph which shows the cost 
of any number of yards of cloth at 12 cents per 
yard. From it we can find the cost of any 
number of yards. For example, the cost of 
15 yd. is found by finding the point on the 
horizontal scale which stands for 15 yd., then 
by finding the point on the cost line directly 
above this point. This appears on the cost line 
as point A, Now, to find the cost of 15 yd. we 
find the point on the cost axis (O Y) horizontally 
opposite the point A which already stands for 
15 yd. The cost proves to be 11.80. Thus 
we see that point A stands both for 15 yd. and 
for $1.80. In the same way the point B shows 
that 8 yd. on the horizontal scale corresponds to 
96 f! on the vertical scale. The point C shows 
that 18 yd. on the horizontal scale corresponds 
to $2.26 on the vertical scale. 

3. Make a formula for the cost of any number of 
yards of cloth at 12 ^ per yard. 



52 Fundamentals of High School Mathematics 

Note that the graph and the formula 
tell exactly the same thing. The graph 
tells the relation between the cost and the 
number of yards purchased more clearly because 
it presents it to the eye as a picture. To tell 
from the graph the cost of any particular 
number of yards requires only a glance ; to tell 
from the formula or equation 

requires that we substitute some particular 
value of 11 in the equation and then that we find 
the value of C. 

4. Draw a graph showing the price of any number 
of pounds of beans at 9 cents a pound. From 
it find the cost of 5| pounds ; of 12 pounds. 

5. Now, write a formula which represents the cost 
of any number of pounds at 9 $ a pound. Note 
that the graph and the formula tell the same thing. 

6. Draw a graph for the cost of a railroad ticket 
at 3 (f, a mile. 

7. If c = .03 ;;/ is used as the equation for the cost 
of any railroad ticket at 3 ^ a mile, show that 
by letting m have particular values, such as 2, 
3, 7, 10, etc., we get values for c } from which 
we can make the graph. 

8. A number of rectangles have the same base, 
5 in. Write an equation for the area of any 
rectangle which has a 5-inch base. (Use h for 
the altitude, or height.) 

9. Draw a graph for the area of any rectangle 
whose base is 5 in. by using the equation you 
got in Example 8. (Let h have particular 



How to Represent Relationship between Quantities 53 

values, such as 2, 3, 4, 7, 10, and find the cor- 
responding area, in each case.) 

10. A west-bound train leaves Chicago at 7 a.m., 
going 30 miles per hour. Show graphically its 
progress until 4 p.m. 

11. Using d = 30 t for the equation of the train in 
Example 10, show that the graph could have 
been made from the results obtained by letting t 
have particular values. 

12. The movement of a train is described by the 
equation d = 25 t. Draw a graph showing the 
same thing. 

13. A boy joined a club which charged an initiation 
fee of 25 cents. His dues were 10 cents each 
month. Draw a graph to show how much he 
had spent at the end of any number of months. 

14. What formula or equation will represent the 
same thing as the graph in Example 13 ? 

VARIABLES AND CONSTANTS 

Section 22. In all the examples which you have just 
solved graphically there have been changing or varying 
quantities ; for example, in the graph of the motion of a 
train, the distance and the time vary as the train moves 
along its trip ; or in any cost graph the cost varies (that is, 
increases and decreases) as the member of articles varies. 

But in these examples, some of the quantities do not 
change or vary. To illustrate : the rate of the train (as in 
Example 10, 30 miles per hour) remains fixed, or constant, 
as the train moves along ; and the price per unit of any 
article (for example, cloth at 12 ^ per yard) remains fixed 
or constant in any particular example. 



54 Fundamentals of High School Mathematics 

Thus, in any problem we may have two kinds of quanti- 
ties : first, those that change or vary ; and second, those 
that remain fixed or constant. We call them, respectively, 
variables and constants. For example, in the formula 
for the area of any rectangle whose base is 4 units, A = 4 //, 
it is clear that A and // are variables, and that the base, 4, 
remains constant. In other words, if h is 2, then A is 8 ; 
if h is 3, then A is 12 ; if // is 7, then A is 28, etc. Thus, h 
can change, but as it changes, A also changes, since A is 
always 4 times as large as //. Hence, 4 is the " constant " 
in the equation, and A and h are the " variables. " Note 
that there is a definite relation between A and h. A is 
always 4 times h. 

EXERCISE 27 

Determine the variables and the constants in each of the 
following examples. Give reasons for each decision that 
you make. 

1. c = 27rR 5. e = 10 m + 25 

2. d = 4Qt 6. A=s 2 

3 A== &A 1 P = 2b + 2k 

4. A'=J + 4 

GRAPHS SHOW THE RELATION BETWEEN TWO VARIABLES 

Section 23. A cost graph, such as Fig. 16, really shows the 
relation between the number of units (lb., doz., or yd., etc.) 
purchased and the total price paid. A graph of the move- 
ment of a train {e.g. Fig. 14) which runs at a constant rate 
shows the relation between the number of hours (the time) 
and the number of miles traveled (the distance). Saying 



How to Represent Relationship between Quantities 55 

that these graphs show the relation between the numbers 
represented by them means that if we read a particular 
value of the time, such as 2 hr. or 5 hr., we can find the 
number of miles which eowesponds to that number of hours. 
Thus, graphs show the relation between two variables; 
that is, they show the values of one variable which corre- 
spond respectively to the values of another related variable. 

A formula also shows the relation or connection between 
the two variables. For example, the formula for the area 
of any rectangle with a 3-inch base, which is A — //, shows 
that tJie value of A must always be three times tJie value of h, 
or, in other words, the area is always three times the height. 
At first it is more difficult to understand the formula than 
the graph, but as you advance in mathematics the formula 
will become more important and significant. 

Thus, as we stated at the beginning of the chapter, there 
are three methods of showing the relationship or connec- 
tion between the kinds of variables we have studied: 

I. THE FORMULA METHOD 
II. THE TABULAR METHOD 
III. THE GRAPHIC METHOD 

Let us illustrate one example by each of these methods. 
A man walks at the rate of 6 miles per hour. Show 
the relation or connection between the distance he 
walks and the number of hours he walks. 

I. Formula method : 

D=Qh. 
II. Tabular method : 

Table 2 



If the no. of hours is 


1 


2 


S 


8 


10 


12 


then the distance is 


6 


12 


30 


4-8 


eo 


72 



56 Fundamentals of High School Mathematics 
III. Graphic method : 





50 


^_ 


j? 


s 


y^ 


40 ~ ; y> _ 




^** 


. » x^ 




30 ■ S 


Jjr ^* 


v3 x 


»* ^^ 


H .?* 


to 20 .^ 


^ x" 


Q ^> 


s 


^ 


10 ^ 


y 


/** 


J? 


s 


S 



3 4 5 

TIME 



8 



Fig. 17. The line shows relationship between time spent and 
distance traveled. 



EXERCISE 28 
PRACTICE IN REPRESENTING THE RELATION BETWEEN VARIABLES 

Show by three methods the relation between the vari- 
ables in the following : 

1. The area of a rectangle whose base is 8 in. and 
its height. 

2. The cost of belonging to a club which charges 
an initiation fee of 50^, and 10^ per month for 
dues. 

3. A freight train leaves Chicago at 10 a.m., at the 
rate of 25 miles per hour; at 1 p.m. a passenger 
train leaves Chicago, running in the same direc- 
tion, at the rate of 40 miles per hour. Show 
graphically at what time the passenger train will 
overtake the freight train. See Fig. 18 for solu- 



How to Represent Relationship between Quantities 57 

tion. How does the graph show that one train 
will overtake the other? If t represents the 















































■ 




250 " 


































































































/ 


















































1 i ),' L 


















































\ 1 V^ 
















































200 " 












































! yP\ 






































' 






' ^ l i 












































^s \ 1 


" 










































<r' 


-.c^t 












































150 






































^^ s> 




































^•s 


l/KI 


































i 1 






































1 !>' 




/ 


































^^ 


/ 






100 " 


































/ 










































/ 








































* 






































I 


/ 








. 
































s 








, 




50 - 
































^ 






















. 














y 






































/ 


■ 1 










































































1 ■ 












« 




















s 















10 



12 



2 3 

TIME 



Fig. 18. The lines show relationship between the time spent and the 
distance traveled by each train. The point of intersection indicates the 
time at which they will meet and how far each travels. 

time of the freight train, what formula will rep- 
resent the distance it travels ? What will repre- 
sent the time the passenger train travels ? 
What formula will represent its distance ? 

4. A slow train left Cleveland at 6 a.m., running 
uniformly at the rate of 30 miles per hour. At 
10 a.m. a faster train left Cleveland, running in 
the same direction, at the rate of 40 miles per 
hour. Show graphically at what time the faster 
train will overtake the slower one. 

5. A freight train left St. Louis at 7 p.m., running 30 
miles per hour. At 11.30 p.m. an express train 
started in the same direction. Show graphically 
at what time it will overtake the freight train, if 
it runs 15 miles per hour. 



58 Fundamentals of High School Mathematics 

Section 24. Two different kinds of graphs. We should 
distinguish between the two kinds of examples which we 
have graphed. The first kind includes all those for which 
no formula or equation can be made. Recall the first 
illustrative example in this chapter: the relation between 
the time of day and the temperature. Clearly, no formula 
can be made which will always show the relation between 
the two variables in this kind of example. Thus, there are 
only two ways to show or represent this kind of relation : 
(1) the tabular method, (2) the graphic method. 

The second kind of example which we have been graph- 
ing is illustrated by any of those examples for which we 
made a formula. For example, we have such illustrations 
as : the graph si towing the relation between the distance 
traveled by a train running at 30 miles per hour and the 
time the train travels. This belongs to the second kind of 
graph, because we can make a formula for the relation 
between its variables. The* formula is : 

d=3Ql. 

Thus, there are three ways to show the relation between 
these variables: (1) the formula or algebraic method, (2) 
the tabular method, and (3) the graphic method. 

In mathematics, we say that the second kind of graph, 
for which an equation can always be made, states alge- 
braic laws, or mathematical laws, because there is 
always a definite relation between the variables. The first 
kind of graph, for which no definite lazv or equation can 
be made, is sometimes called a statistical graph. It is 
this kind that is most frequently seen in newspapers and 
magazines. In mathematics, however, the other kind, that 
which states " laws," is nearly always used. 



How to Represent Relationship between Quantities 59 

The next exercise will give practice in making both 
kinds of graphs. It is important to tell whether the in- 
formation to be graphed (generally called the data) can be 
expressed by an algebraic law or formula. 



EXERCISE 29 

l. The following table shows the average heights 
of boys of different ages. Construct a graph 
showing this information or data. 



Age in years 


2 


4 


6 


8 


IO 


12 


14 


16 


ia 


20 


Height in feet 


1.6 


2.6 


3.0 


3.5 


4.0 


4.8 


5.2 


5.5 


5.6 


57 



3. 



5. 



6. 



Represent ages on the horizontal scale. 
When does the average boy grow the most 
rapidly ? the most slowly ? 

Is there an algebraic " law," or formula, which 
shows the relation between these two variables, 
age and height ? 

Mr. Smith joined a lodge which charged 825 
initiation fee, and dues of $2 per month. Show 
graphically the relation between the cost of be- 
longing and the time one belongs. 
Is there an algebraic "law," or formula, which 
shows the relation between the variables, cost 
and time ? 

The information or data of the following table 
represent the area of a square of varying sides : 



If the side is 


1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


then the area is 


1 


4 


9 


16 


25 


36 


49 


64 


81 


100 



oo Fundamentals of High School Mathematics 

Show this relation between the area of the 
square and its side graphically, using the verti- 
cal scale to measure areas and the horizontal 
scale to measure sides. 
7. Is there an algebraic "law," or formula, which 
shows the relation between the variables here ? 

Section 25. Summary of chapter. This chapter should 
make clear the following truths : 

1. Important facts about quantities are more easily- 
read and interpreted if they are represented 
graphically. 

2. Graphs always show the relation between two 
varying quantities. 

3. Two scales, a horizontal scale and a vertical scale, 
at right angles to each other, are required in order 
to mark off or measure the values of the varying 
quantities. These scales must be divided into 
convenient units. 

4. The information or data must be tabulated in order 
to show it graphically. 

5. There are three fundamental methods of describ- 
ing the relationship between related variables : 

a. The Formula, or Algebraic Method, of stating 
"Law" ; 

b. The Tabular Method of expressing " Law" ; 

c. The Graphic Method of expressing " Law." 

EXERCISE 30 

l. If n represents a boy's present age, state in 
words what the expression n + 7 = 22 means. 



How to Represent Relationship between Quantities 61 

2. Give a formula for the base of a rectangle when 
the area and height are known. 

3. Represent the number of cubic yards in a box- 
shaped excavation when the dimensions are ex- 
pressed in feet. 

4. If m, s, and d represent the minuend, subtra- 
hend, and difference respectively, what formula 
will show the relation between these numbers? 

5. Show by a formula the relation between the 
product,/, multiplicand, M, and multiplier, vi. 

6. Give the meaning of the formula i=prt. 

7. Divide each side of the formula V= Iwh by Iw 
and tell what the resulting formula means. 

8. Give a formula for the volume of a cube whose 
edge is s. 

9. Evaluate the above formula when s = 3.2. 

10. Translate into words the formula d = rt. 

11. Divide each side of the formula d— rt hy r and 
tell what the resulting formula means, 

12. In the formula c = np, n represents the num- 
ber of articles bought, p represents the price 
of each, and c represents the total cost. Trans- 
late it into a word statement. 

13. Divide each side of the formula c = np by n, 
and tell what the resulting formula means. 

14. Does x = 4t satisfy the equation x 1 + 6 x = 40 ? 

15. Solve the equation 10 j/ + 7 = 52 + 4 y. 

16. Is the equation x +y + 3 = 20 satisfied if x = 8 
and y = 9 ? Can you find any other values of 
x and y which will satisfy this equation ? 



02 Fundamentals of High School Mathematics 



(*) 



17. Solve each of the following equations, thinking 
of each example as asking a question : 

2^ = 5.5 

y 

(b) .5/ =17 (/) 1.5* = 45 

(<:) .4/ = 80 Qr) f 2 = 60 

34 ,„r ... ' h 



{a) ~- = 2 
x 



V) 



4.L>f> 



(//) 



.34 



= 85 



Tell what you do to each side of the equation ; 
that is, tell whether you add, subtract, multiply, 
or divide, on each side. 



REVIEW EXERCISE 31 

1. Write in algebraic language : The volume of a 
sphere is four thirds the cube of the radius 
times 7r. 

2. The weights of a baby boy who weighed 8 lb. 
at birth are given for each month of his first 
year by the table : 



Month 


1 


2 


3 


4 


5 


6 


7 


8. 


3 


io 


11 


12 


Weight 


9* 


11* 


I2f 


14* 


151 


I65 


18 


19 


19| 


20 


21 


22 



Represent this graphically. 

3. Which would you rather have, ix + by dollars, 
or 5 x 4- 4 y dollars, if x = % 20 and y = $ 1 6 ? 

4. If V is the volume of a cone, b the area of its 
base, and // its height, then 

V= \bh. 

Write this formula in words. 



How to Represent Relationship between Quantities 63 

5. Nurses keep temperature records of fever pa- 
tients. For one patient the following degrees 
of fever were noted : 

2, 5.4, 4, 6.1, 4.5, 6.5, 5.3, 6.7, 4.5, 6,5, 5.9, 6.2, 
6.9, 5, 6.4, 4.7, 5.8, 7.6. 

These readings were taken an hour apart. 
Show graphically this patient's successive tem- 
peratures. 

6. If you know that 

10a - 7 = ±a + 35, 
then what do you do to each side to get 

10^ = 4^ + 42? 
How do you get 6 a = 42 ? 
Why does a = 7 ? 

Prove that a = 7. 

7. Get the hourly temperature for 24 hours from 
your daily paper, and construct a graph to rep- 
resent the changes in temperature. 

8. The sum of two numbers is 18 and their differ- 
ence is 4. What are the numbers ? 

9. Show that 2 a and a 2 are unequal by choosing 
some particular value for a, such as a = 6. Do 
you think there is any possible value for a 
which would make 2a = a 2 ? 

10. What product is obtained by using. 7 as a 
factor twice ? by using 2 a as a factor three 
times ? 



CHAPTER V 

HOW TO FIND UNKNOWN DISTANCES BY MEANS OF 
SCALE DRAWINGS: THE FIRST METHOD 

Section 26. We need to know how to find unknown dis- 
tances. The methods of mathematics are really all planned 
to help us find unknown values. The equation, which we 
have studied so carefully, is the best algebraic tool with 
which to do that. Many times, however, in practical life 
work the unknown values that we need to know are dis- 
tances. For example, the surveyor may need to know the 
distance across a river and may not be able actually to 
measure it. Or, he may need to know the distance be- 
tween two points, with some other intervening object 
between which prevents him from measuring it directly. 
Now, mathematics has given us three ways to find such an 
unknown distance. In Chapters V, VI, and VII we shall 
discuss these methods. 

The first method is to make a scale drawing, which will 
include in some way the unknown distance. Next, there- 
fore, we shall study how to determine unknown distances by 
means of scale drawings. Before we take up that particular 
subject, however, we must study. how to measure the lines 
and angles which make up scale drawings. 

Section 27. The measurement of lines. We are already 
familiar with certain methods of measuring distances. For 
example, we have measured the length of lines, such as the 
distance from A to B or from C to D. If we use a metric 
scale, in which the units are centimeters, the distance from 

A L 

C D^ 

6 4 



Finding Unknown Distances by Scale Drawings 65 

A to B y which is read "line AB," is 5.08 centimeters long, 
and the line CD is 6.35 centimeters long. If we use a foot- 
rule in which the units are inches, the distance between A 
and B y or the line AB, is 2 inches, and line CD is 2.5 
inches. Note here that the distances or lengths that we 
obtain for these lines depend upon the kind of scale, or 
kind of unity that is used in measuring. 

THE MEASUREMENT OF ANGLES 

Section 28. An angle is determined by one line turning 
about another. In order to construct scale drawings, we 
must know how to measure angles. Let us think of an 
angle as being formed by one line turning, or rotating 
about a fixed point on some fixed or stationary line. The 
line O Y turns or rotates about point O. For example, in 




Fig. 19 



Fig. 19, think of AX as a fixed, or stationary, line. (It is 
easiest always to take this line as horizontal.) Think also 
of another line, say O Y y as turning, or rotating, about some 
point on the fixed line AX 9 say point O. As the line OY 
rotates about the point O f it constantly forms a larger and 
larger angle with the fixed line AX. (The symbol for 
angle is Z.) The point O, about which the line turns, is 
always the point at which the two sides of the angle meet, 
and is called the vertex of the angle. 



66 Fundamentals of High School Mathematics 

The arrow is drawn to indicate that the line OY is turn- 
ing, or rotating, about the point 0. 

Section 29. The unit of angular measurement. Just as 
we have units and scales for measuring straight lines, so 
we have units and scales for measuring angles. Evidently 
the unit with which we must measure the size of the angle 
is one that will measure the amount that the line has rotated 
about the fixed point. Figure 20 shows that we can think 
of the rotating line as turning clear around until it occupies 
its original position again. That is, any point P on 
the line OX has 
turned through a 
complete circle in 
rotating about O 
and returning to 
its original posi- 
tion. 

This suggests 
that the unit with 

i • i Fig. 20 

which we measure 
angles zvill be some definite fraction of the circle. Fdr a 
long time people have agreed that the circle be divided 
into 360 units and that each one of these units of angular 
measure be called a degree. The symbol used for 
degree is a small ° placed at the right above the number. 
For example: 45° is read "45 degrees." Thus, Figs. 21, 
22, and 23 illustrate angles of different sizes or of differ- 
ent numbers of degrees. 




Finding Unknown Distances by Scale Drawings 67 





Fig. 21 



Fig. 22 



s 135 




Fig. 23 



Section 30. The PROTRACTOR: How to measure 
angles. Just as we use foot rules, yardsticks, meter 
sticks, etc., to measure straight-line distances, so we have 
an instrument called a protractor to measure angular 
distances. Figure 24 shows that the circular edge of the 




Fig. 24. A protractor for constructing and measuring angles. 



6S Fundamentals of High School Mathematics 

protractor is marked off (*.*■ is " graduated ") into degrees. 
Note from the figure that the protractor is. divided into 
180 equal parts (half of the total number of angular units 
in the circle), called degrees. Sometimes the whole circle 
is used and marked off to give 300°. 




Fig. 25 



The next figure, Fig. 25, shows how to measure an angle 
with a protractor. First, lay the straight edge of the pro- 
tractor so that it will fall exactly upon one of the two 
lines that form the angle, and with the center of the pro- 
tractor exactly upon the vertex, O, of the angle. Then 
the other side of the angle, OB y for example, will appear 
to cut across the circular edge of the protractor. Now 
count the number of degrees from the point where the 
curved edge of the protractor touches OA to the point 
where it crosses the line OB. Hence, in Fig. 25, the 
angle A OB contains 54°. It is very important for us to 
be able to read angles accurately. The next exercise will 
give you practice in reading angles. 



Finding Unknown Distances by Scale Drawings 69 



EXERCISE 32 
PRACTICE IN MEASURING ANGLES 

1. Measure each of these angles with a protractor. 

B 




Fig. 26 



Fig. 27 




Fig. 28 

Compare angle A and angle C. Which has the 
longer sides ? What effect has the length of a 
side of an angle upon the size of the angle ? 
Measure each angle of triangle ABC. From the 
results of your measurement, what is the sum 
of all three angles of this triangle ? 




Fig. 29 



jo Fundamentals of High School Mathematics 



How large is Z.r? 
How many de- 

i n Z y ? 

? How 



X 



g r e e s 

in Z; 

many 



degrees 




Fig. 30 



in the sum of the 
angles of this tri- 
angle, XYZ? 

5. Draw with the protractor an angle of 30° ; 45° ; 
60°; 100°. 

6. At each end of a line 6 cm. long draw angles of 
50°. Produce these lines until they meet, and 
measure the angle formed by them. How many 
degrees in it? Compare the lengths of the lines 
you drew. How many degrees does the sum of 
the three angles of this triangle make ? 

7. Draw a triangle 
such as triangle 
ABC, so that AB 
= 4 inches, angle 
^ = 60° and AC 
= 3 inches. Then 
find the number , ._. 
of degrees in an- ■**■ )<- 
gle B and angle C 

8. Construct triangle ABC so that 
AC=r> cm., angle C=40°, and 
CB = 5 cm. Compare angle A 
with angle B. How many de- 
grees in each ? 

Explanation : A triangle having two 
sides equal, such as AC and CB, is an 
isosceles triangle. It is proved in geom- Fig. 32 




Fig. 31 




Finding Unknown Distances by Scale Drawings 71 



etry that the angles opposite these equal sides are always 
equal; for example, angle A = angle B. How many degrees 
ought there to be in either angle A or angle B ? 

Section 31. How to describe an angle. An angle is 
described by using three letters, i.e. the letter which repre- 
sents the vertex is written 
between the two letters at 
the ends of the sides. Thus, 
Z 1, in Fig. 33, is read as 
angle A OB or angle BOA, 
and is written Z A OB or 
Z BOA. In the same way, 
Z 2 is read angle BOC or 
angle COB, and is written 
Z BOC or Z COB. 




Fig. 33 



EXERCISE 33 
PRACTICE IN READING ANGLES 

1. Why would it not be clear to read Z2 as Z O? 

2. Read the angle formed by lines OA and OB. 

3. Read the angle formed by lines OB and OC. 

4. Determine the number of degrees in ZAOB, in 
Fig. 34, without using the protractor* 





Fig. 35 



72 Fundamentals of High School Mathematics 

5. If in Fig. 35 you 
know that angle 
ABC is 40° and 
that Z BCA is ( <>0°, 
could you find 
Z CAB without 
measuring it? 
How ? How large is it ? 

Section 32. We must be able to find unknown distances 
which cannot be measured directly. The preceding sec- 
tion took up only examples in which the distances, linear 
and angular, could be measured directly, by means of 
instruments. There are many instances, however, in 
which the lengths of the lines and the sizes of the angles 
cannot be measured directly. For example, consider the 
case of finding the distance across a river, or the height of 
a tree, which we mentioned at the beginning of the chapter. 
In cases like this we need indirect methods of measuring. 
Mathematics makes it possible for us to determine the 
lengths of such lines by measuring the lengths of other 
lines and the sizes of angles that are related to them. This 
leads us to the main topic of this chapter. 

HOW TO FIND UNKNOWN DISTANCES BY MEANS OF 
SCALE DRAWINGS 

Section 33. How to draw distances to scale. One of the 

methods that you will use commonly in indirect measure- 
ment is that of drawing distances "to scale." So much 
use is made of mechanical drawings that we need to be 
very proficient in making them and in reading them. Let 
us take a simple illustration of the drawing of distances 
"to scale " and of measuring distances on scale drawings. 



Finding Unknown Distances by Scale Drawings 73 

Illustrative example. A man starts at a given point and 
walks 2.5 miles east, then 2.5 miles north. How far is he from 
his starting point ? 

First, set point 0, in Fig. 36, as his starting point. East is meas- 
ured to the right of point O and north above point O. 
Second, to represent distances "to scale," we need to select a 































b\ 


c 
























•ft 




























x? 


>/ 
























<- 


























y 












6 














, n c 


% 












in 












x! 


^ 














(N 










<^ 


^ 
















,rj 








xf 


V 


















*£3 






«? 


^ 

























, ll 


§ 




























5) 
























1 




•*/ 


























f 


k 






Ej 


as 


t . 


2. 


5i 


iii 


let 






— > 


F 


> 



Fig. 36 



wniY 0/ distance on the scale which will represent a unit of distance 
in the example. Let us take, for example, three quarters of an 
inch on the drawing to represent each mile which the man 
actually walks. This is indicated on the scale drawing (Fig. 36) 
by writing "Scale = § in. to 1 mi." It is very important to 
select the scale unit carefully and always to indicate the scale 
that has been used on the drawing. 

Third, to represent the man's path, we lay off OB horizontally 
to the right of O, 2.5 miles (on the drawing this amounts to 
If inches) and BC vertically, 2.25 miles. Then, by using the 
cross-section paper as a scale, we can measure at once the distance, 
OC, that the man is from his starting point. The distance is 
2.63 inches on the figure, or 3.54 miles actually. 



74 Fundamentals of High School Mathematics 

This work illustrates by a very simple example how we 
make scale drawings. Mechanical drawings made " to 
scale " are used very commonly by such workers as archi- 
tects, carpenters, machinists, and engineers. 



EXERCISE 34 

PRACTICE IN FINDING UNKNOWN DISTANCES BY THE CONSTRUCTION OF 

SCALE DRAWINGS 

1. Draw to the scale 1 cm. to 2 ft. a floor plan of 
a room 28 ft. by 20 ft. By measuring the 
distance diagonally across the plan, compute 
the diagonal of the room. 

2. Draw a plan of a baseball diamond 90 ft. square 
and find the distance from first base to third 
base. Use 1 cm. to represent 20 ft. 

3. Two bicyclists start from the same point. One 
rides 12 miles north and then 8 miles east; 
the other rides 10 miles south and then 
6 miles west. How 

far apart will they be ? 
Use the scale 1 cm. to 
2 mi. 

4. Draw to the scale 1 cm. 
to -4 ft. a plan of the 
end of a garage such 
as in Fig. 37. Find 
the height from the 
floor to the top of the 
roof. Fig, 37 







Finding Unknown Distances by Scale Drawings 75 

5. In Fig. 38 AB is 
200 ft., angle A is 
60°, and angle B 
is 50°. Draw to the 
scale 1 cm. to 50 ft. 
and determine the 
length of AC and 
BC. 

6. A surveyor sometimes finds it necessary to 
measure the distance across a swamp, such as 




Fig. 38 




Fig. 39 



AB in Fig. 39. He measures from a stake 
i4toa stake C, 120 ft. From C to B he finds 
it is 100 ft. Find, by a scale drawing, the dis- 
tance AB across the swamp, if angle C is 85°. 

How could a surveyor find the distance from 
A to B, if there were some obstacle in the way 



j6 Fundamentals of High School Mathematics 






«^> 




A 



Fig. 40 



B 



preventing his measuring directly the distance 
AB? 

8. Find by a scale drawing the distance AC across 




Fig. 41 



the river, if it is known that angle A = 80°, 
AB = 200 ft., and angle B = 70°. 

9. Illustrative example. A boy wishes to determine the 
height of a flagpole. A scale drawing will aid him in 



j 


l q 




r; 


\ 


s 


\ 


O 


\ 


6 




P 


^ \ 






known L f 1 \ 


! 


, B 


l \ known Z.\ rp 






<— —80' Known A 



Fig. 42 



Finding Unknown Distances by Scale Drawings 77 

doing this. For example, he can measure a line of any 
length on the ground out from the base of the flagpole. 
Suppose he takes a line 80 feet long. Then he sets at E an 
instrument called a transit, with which he can read the 
angle between the horizontal base line and the line of sight 
from E, where he stands, to H, the top of the pole. He 
knows also that the angle B is a right angle. So he 
knows the length of the line EB, the size of the angle E and 
the angle B. He constructs a scale drawing to represent 
the known lengths and the known angles. From this 
drawing he is able to " scale " or measure the height of the 
flagpole. 



The angle BEH or angle E between the horizontal and 
the line of sight in this example is called the angle of 
elevation. If the boy had taken a longer base line, what 
would have been true of the size of the angle of elevation 
with respect to what it was before ? 

10. If the angle of elevation in Fig. 42 is 40° when 
the observer is 80 ft. from the foot of the pole, 
find its height. 

11. A flagpole 50 ft. high casts a shadow 60 ft. 
long on level ground. What is the angle of 
elevation of the sun ? If the length of a shadow 
cast by this pole increases, what conclusion can 
be drawn concerning the angle of elevation ? 

12. The angle of elevation of the top of a tree is 
42° when the observer stands 30 yd. from the 
tree. How high is the tree ? If the distance 
from the observer to the tree decreases, what 
change in the angle of elevation follows ? 



78 Fundamentals of High School Mathematics 

13. On the top of a church is a tall spire. At a 
point P on level ground 75 ft. from the point D 
directly beneath the spire, the angle of eleva- 
tion of the top of the spire is 38°. At a point 
Q 150 ft. from D, in line with P and D, 
the angle of elevation of the top of the spire 
is 28°. Find the distance from Q to the top of 
the spire. 

14. An anchored balloonist from a height HT, 
Fig. 43, of 2500 yd., observes the enemy at D. 



BALLOON 




Fig. 43 



He wishes to compute the distance DT, on 
level ground. To do so, he measures the angle 
which is formed by the horizontal line HL and 
the line of sight HD. This angle is called the 
angle of depression, and has the same num- 
ber of degrees as the angle of elevation. (Can 
you see that this is true ?) He next finds angle 
THD by subtracting the angle of depression 
from 90°, angle THL. He then knows enough 



Finding Unknown Distances by Scale Drawings 79 

about the triangle DTH to make a scale draw- 
ing of it. Find DT if the angle of depression 
is 35°. 

15. From the top of a lighthouse 80 ft. high the 
angle of depression of a ship is 35°. How far 
is the ship from the base of the lighthouse? 
Compare your result with that obtained by the 
other members of the class. 

16. From the top of a cliff 120 ft. above the sur- 
face of the water the angle of depression of a 
boat is 20°. How far is it from the top of the 
cliff to the boat ? 

17. A searchlight on the top of a building is 180 
ft. above the street level. Through how many 
degrees from the horizontal must its beam of 
light be depressed so that it may shine directly 
on an object 400 ft. down the street from the 
base of the building ? How does your result 
compare with that obtained by other pupils ? 

18. From the tenth story of a building the angle 
of depression of an object on the street level is 
30° ; from the eighteenth story the angle of de- 
pression of the same object is 50°. Find the 
distance from the object to the base of the build- 
ing if the second observation point is 80 ft. 
above the first observation point. 

19. An observer is 200 ft. from the ground. The 
angle of depression of a point A is 24°, of a 
point B 42°, and of a point C 15°. Which 
point is closest to the observer ? farthest from 
the observer ? 




8o Fundamentals of High School Mathematics 

20. In Fig. 44 measure (1) the angle of elevation 
of point D from point E ; (2) the angle of de- 
pression of point E from 
point D. Compare these 
angles. Note that line 
DH is parallel to EC, 
and that these angles are 
formed by line DE cut- 
ting (or intersecting) 
these two parallels. In 
geometry it is proved 
that such angles are al- 
ways equal. 

21. Is the scale drawing a very accurate method of 
determining unknown distances ? Do many of 
the pupils in the class get the same answer for 
any particular example ? Why ? 

A NEW WAY TO COMPARE TWO QUANTITIES; NAMELY, 
TO FIND THEIR RATIO 

Section 34. We have been comparing two quantities by 
finding how much larger or smaller one quantity was than 
another quantity. For example, if the line AB is 4 units 
long and the line CD is 6 units long, we should have said, 
in describing the comparative lengths, that the line AB 
was 2 units shorter than line CD, or that line CD was 
2 units longer than the line AB. 

A B 

C D 



Another method, however, of comparing quantities is 
used very extensively in mathematics. It is the method of 
dividing one quantity by the other, or finding the quotient 



Finding Unknown Distances by Scale Drawings 81 

of the two quantities. Thus, to compare the line AB with 
the line CD we divide AB by CD, which gives : 

AB = 4: 

CD 6* 

This result is read " the quotient, or ratio, of AB to CD 
is equal to four sixths." This process is described as find- 
ing the ratio of the two lines. The ratio of two numbers 
means, then, the quotient which results from dividing one 
of the numbers by the other. Thus, the ratio of 5 to 10 is 
|-, and is written T 5 ^ = \. The ratio of 10 to 5 is 2, and is 
written -5°-= 2. In the same way the ratio of 1 in. to 1 ft. 
is -^ ; the ratio of \ to | is f ; and the ratio of ^x to 1 x is 

4:X 4 

7 x 7 

EXERCISE 35 
PRACTICE IN DEALING WITH RATIOS OF NUMBERS 

1. What is the ratio (in lowest terms) of 10 to 12 ? 
of 20 to 24 ? of 15 to 8 ? of 25 to 30 ? Show 
that 5x and Qx represent all pairs of numbers 
whose ratio is f . 

2. Give several pairs of numbers having the ratio f . 
Show that 3x and 4.r represent all pairs of num- 
bers having this ratio. 

3. The ratio of two numbers, a and b, is \ . What 
is b when a is 40 ? 

4. The ratio of two lines, m and ;z, m 

|. Find the length of m if 71 n 



is 18 in. 

5. Divide 40 into two parts whose ratio is f . 

6. A father and his son agreed to divide the profits 
from their garden in the ratio of |. Find each 
one's share if the total profits were $210. 



8j Fundamentals of High School Mathematics 

7. The ratio of Z A to Z. B is f. Find Z B when 
ZAis 80°. 




Fig. 45 



8. The ratio of the areas of the two squares, S x and 
S 2f is \. Find the area of each if the sum of 
their areas is 45 sq. in. 



10. 





Fig. 46 



Divide an angle of 90° into two angles having 
the ratio of 4 to 5. 

Measure each angle, Fig. 47, with a protractor. 
Find their ratio. 





Fig. 47 



Finding Unknown Distances by Scale Drawings 83 

11. In the two triangles ABC and XYZ, what is the 
ratio of Z CtoZZ? Of ZAto ZX? Oi Z B 
to Z Y? Do these triangles have the same 
shape ? Do all triangles have the same shape ? 





Fig. 48 

12. Find the ratio of two lines if one is 2 feet long 
and the other 3 yards long. 

13. What is the ratio of 3 pints to 4 quarts ? 

14. 2 = A. 15 . 4.5. 16 . 5 10. 

n 15 3 6 In 

17. Find the ratio of AB to CD, Fig. 49, by measur- 
ing the length of each line. Express the result 
decimally. 

A B 



Fig. 49 

18. A school baseball team won 7 of the 10 games 
it played. What was its standing in percentage. ? 

Solution : The standing of the team in percentage means 
the ratio of the number of games won to the number played, 
expressed decimally. Thus, the standing of this team is ex- 
pressed by the result obtained from changing the ratio, T 7 o, 
to a decimal. This gives .70 or .700. 



84 Fundamentals of High School Mathematics 

19. What was the percentage or standing of a team 
which won 9 of its 12 games ? 

20. The winning team in one league won 14 of its 
18 games ; and the winning team in another 
league won 15 of the 19 games it played. Which 
had the higher percentage ? 

21. Express each of the following ratios as a deci- 
mal, correct to two places : 



/ \ 3 


«>1 


( \ 5 
{e) 16 


( ^| 


... 22 
W 81 





«.£ 


C/)f 


(/ ' } 45 


, ., 16.2 
{J) 21 



22. If 12 quarts of water are added to 25 gallons 
of alcohol, what is the ratio of the water to the 
entire mixture ? Express decimally. 



REVIEW EXERCISE 36 



l. A boy knows that AB is 100 ft. and that 
Z. B = 40°. From this information can he con- 
struct a scale drawing for the triangle ? Give 
reasons for your answer. 



100 ft 




2. What facts or data must be known about a tri- 
angle before you can make a scale drawing of it ? 



Finding Unknown Distances by Scale Drawings 85 

3. How does the scale of a drawing illustrate ratio ? 

4. A tree 90 ft. high casts a shadow 140 ft. long. 
Find from a scale drawing the angle of elevation 
of the sun. 

5. The following table shows the humber of feet 
required to stop an automobile running at vari- 
ous speeds. 



At a speed of 
(miles per hour) 


IO 


]5 


20 


25 


30 


35 


40 


50 


a car should 
stop (feet) 


9.2 


20.8 


37 


5Q. 


83.3 


104. 


148. 


231. 



Represent this graphically, measuring the speed 
on the horizontal axis. 

6. Represent in the briefest way the product of 
five x's ; the sum of five x's. 

7. Does a 2 b = ab 2 if a is 4 and b is 3 ? 

8. Can you construct a triangle similar to Fig. 50 
if you know that AB is 200 ft. and AB = 40°? 
Why? 

9. What must be known about a triangle before 
you can construct it accurately ? 



CHAPTER VI 



A SECOND METHOD OF FINDING UNKNOWN 
DISTANCES: THE USE OF SIMILAR TRIANGLES 

Section 35. Scale drawings are very inaccurate. In the 
last chapter we saw that scale drawings could be used to 
find unknown distances, either linear or angular. The 
results obtained, however, were very inaccurate. Seldom 
did many of you get the same answer for any example. 
Therefore we need more accurate metliods for determining 
unknown lines and unknown angles. This chapter, and 
the next one, will show methods that depend less upon the 
accuracy or skill of the person who "scales" or measures. 
The first method, based ?tpon geometrical figures of exactly 
the same shape, will be explained now. 

Section 36. What are similar figures? You have al- 
ready seen many objects or figures of exactly the same 
shape. A scale drawing has the same shape as the figure 
from which it was made ; on a photographic plate the figure 
is the same in outline or shape as the original ; the map of 
a state has the same shape or outline as the state itself. 

Figures which have the same shape are said to be simi- 
lar figures. Which of the following figures are similar 
in shape ? 




o 



Fig. 53 



Fig. 54 




Fig. 55 




Fig. 51 



Fig. 52 



Fig. 56 



86 



Finding Unknown Distances by Similar Triangles 87 

Similarity in shape in geometrical figures is a very im- 
portant principle that we are able to use in many ways in 
mathematics. Before this can be taken up, however, we 
need to be perfectly clear as to what is meant by similarity 
of figures. The next exercise will help to do this. 

EXERCISE 37 
PRACTICE WITH SIMILAR TRIANGLES 

1. Draw a line XY twice as long as AB, Fig. 57. 
At X draw an angle equal to angle A. At Y 
draw an angle equal 
to angle B. Produce 
the sides of these 
angles until they meet 
at Z. Measure the 
angle formed by these 
sides. How should it 
compare with angle C? 
Why? 

2. (a) Angle A corre- 
sponds to what angle in your triangle ? 
(b) Angle B corresponds to what angle in your 
triangle ? 

{c) What is true, then, about the corresponding 
angles of the two triangles ? 

3. Measure the side in your triangle which corre- 
sponds to AC, and the side which corresponds to 
BC. What is the ratio of AB to XY, or what is 

the value of ? of ? of ? What does 

X Y X/l, Y Z, 

this tell about the ratios of corresponding sides ? 




Fig. 57 



88 Fundamentals of High School Mathematics 



Construct a tri- 
angle larger than 
Fig. 58 but having 
its angles equal to 
the angles of Fig. 
58. Is your tri- 
angle the same 
shape as Fig. 57 ? 




1.5" 



Fig. 68 

After careful measurement find the ratio of AB 
to its corresponding side in your triangle. Then 
find the ratio of AC to its corresponding side. 
Compare these ratios. 





Fig. 59 



Fig. 60 



5. In Figs. 59 and 60 it is known that 



AC 



2 

3' 



BC 2 , AB 2 



XZ 

In other words, it is 



known that the ratios of corresponding sides are 
equal. Measure each angle of each triangle and 
compare each angle in Fig. 59 with its corre- 
sponding angle in Fig. GO. State carefully the 
conclusion to be drawn from this comparison. 
Are these triangles of the same shape ? 

Section 37. Similar triangles. The previous exercises 
illustrate two very important and widely used truths about 
similar triangles : 



Finding Unknown Distances by Similar Triangles 89 

(1) IF IT IS KNOWN THAT THE ANGLES OF ONE TRIANGLE ARE EQUAL 
RESPECTIVELY TO THE ANGLES OF ANOTHER TRIANGLE, IT FOLLOWS THAT 
THE RATIOS OF THE CORRESPONDING SIDES OF THE TRIANGLES ARE EQUAL. 

(2) IF IT IS KNOWN THAT THE RATIOS OF THE CORRESPONDING SIDES 
OF TWO TRIANGLES ARE EQUAL, IT FOLLOWS THAT THE ANGLES OF ONE 
TRIANGLE ARE EQUAL RESPECTIVELY TO THE ANGLES OF THE OTHER 
TRIANGLE. 

Thus, to state that two triangles are similar is equiva- 
lent to stating that the corresponding angles are equal \ and 
that the ratios of the corresponding sides are equal. 

This principle or truth is used very much in mathe- 
matics. To illustrate, suppose we know that the angles of 
one triangle are equal respectively to the angles of another 
triangle ; then we also know that the ratios of correspond- 
ing sides are equal. Hence, we can make an equation from 
these equal ratios and from this equation find important 
unknown distances. The next exercise will show how this 
is applied to finding the length of lines. 

EXERCISE 38 



ADDITIONAL PRACTICE WITH SIMILAR FIGURES 

1. Figure 61 is a 
right triangle. 
Why? If angle 
A is 30°, find an- 
gle C. If angle 
A is one fourth 
of angle C, find 
the size of each 
angle. 

2. In a right triangle one of the acute angles (that 
is, one of the angles smaller than a right angle) 
is 40°. Find the other acute angle. 




Fig. 61 



90 Fundamentals of High School Mathematics 




Fig. 62 



Fig. (>3 



3. 



4. 



6. 



Figures 62 and 63 are right triangles. If angle 

A = angle D, are the triangles similar ? Why ? 

In Figs. 62 and 63, if EF=6, ED = 8, 

and CB=3, what must AB equal? To solve 

this problem we use the principle that the ratios 

of the corresponding sides of similar triangles 

x 3 
equation q=77- 

8 D 



are 



equal. This gives the 

What, therefore, is AB ? 

The sides of a triangular plat of ground are 
150 ft, 100 ft, and 125 ft, respectively. The 
side of a scale drawing of this plat, correspond- 
ing to the 150-foot side, is 5 cm. Find the side 
of the scale drawing corresponding to the 
100-foot side. Solve as in Example 4. 
The sides of a triangle are 3, 4, and 5 cm. The 
shortest side of a similar triangle is 16 cm. 
Find the other sides of the second triangle. 
A house is 36 ft. high and the garage is 16 ft. 
high. If the house is represented in a drawing 
as 18 in. high, how high should the drawing 



Finding Unknown Distances by Similar Triangles 91 

of the garage be ? What mathematical prin- 
ciple is used to show this ? 

8. Two rectangular gardens are the same shape, 
but of different size. The larger one is 72 ft. 
by 84 ft. If the length of the smaller one is 
40 ft., what must be its width ? 

9. Two angles of one triangle are equal respec- 
tively to two angles of another triangle. Are 
the triangles similar ? Why ? 

10. Line AB is parallel to line CD. Would they 
meet if produced, either to the right or to 
the left of 
the third line 
MM? Meas- 
ure Z 1 and 
Z 2. These 
angles are 
called cor- }sj' 
responding fig. 64 a 

angles of parallel lines. 

11. In Fig. 64 measure the other pair of correspond- 
ing angles, Z 3 and Z 4. What do you find ? 

These two exercises illustrate a very important fact in 
mathematics ; namely, that the corresponding angles of 
parallel lines are always equal. Later on this will be 
proved without measuring the angles ; that is, without any 
possibility of error. You will make use of this fact with- 
out again measuring the angles. £ 

12. In triangle ABC, DE is drawn />v 
parallel to AB. DoesZl = Z2? 1^4- -_XE 

Why ? Is triangle DEC similar j\^ -^B 

to triangle ABC? Why ? fig. 64 b 




92 Fundamentals of High School Mathematics 



13. 



14. 



15. 



16. 



In Example 12, DC= 12, AC = 21, and CE = 10. 
Show how BC can be found, by using the prin- 
ciple that the ratios of corresponding sides of 
similar triangles are equal. What is the length 
of BC? 

A boy wishes to measure the height of a tree. 
He notes that the tree, AC, its shadow, AB y 

c 




J 
2« 



STICK 




k-10'— I 



Fig. 65 



and the sun's ray, CB, passing over the top of 
the tree, form a triangle. He measures the 
shadow and finds it 100 ft. long. At the same 
time a vertical stick 4 ft. high makes a shadow 
10 ft. long. Why is the triangle formed by the 
stick, its shadow, and the sun's ray passing over 
the top of the stick similar to the other triangle ? 
How can the boy find the height of the tree 
from the similar triangles ? What is its height ? 
A Boy Scout wagered he could find the distance 
between two trees, A and B, on opposite sides 
of a river, without crossing it. Could he do it, 
and if so, how ? If not, why not ? 
A crude way to measure the height of an ob- 
ject is by means of a mirror. Place a mirror 



Finding Unknown Distances by Similar Triangles 93 

horizontally on the ground at M y and stand at 
the point at which the image of the top of the 




Fig. 66 

object is just visible in the mirror. Show how, 
by measuring certain distances, this would 
enable one to compute the height of the object. 

17. In triangle ABC, DE is parallel to CB. Show 
that triangle BED is similar to triangle ABC. 
If BC= 10, ED = 5, and AE = 8, what is AB ? 

18. Show that in Fig. 67 the ratio of DE to AE, or 
DE 



AE 



, remains the same even if DE is drawn in 



different positions (always parallel to CB). 



n 








O 


\v. D 




M 


r 






V 


_L 






B 


E 
* ? - 


« £ 




^ 



Fig. 67 



94 



20. 



Fundamentals of Higli School Mathematics 

% Y 



19. 




Fig. (i8 

Figure 08 shows two triangles, with the size of 
each angle indicated, which a teacher drew 
upon the blackboard for an examination. She 
asked the following questions about the two 
triangles : 
(a) Are they similar triangles ? 

AB _AC 

ZY 

AC 



Why? 



{b) Does 

(c) Does 

(d) Does 



XY 
ZY_ 

AB~~ 



XY 

_BC ? 

' ZY' 

XZ , 

: BC 



Why? 
Why? 
Why? 



(<r) Does the ratio of any two sides equal the 
ratio of any other two sides ? 
How would you have answered these questions ? 
The sides of a small triangle are 3, 4, and 6. 
Is it similar to a larger triangle whose sides are 
15, 18, and 80 ? See Section 37. 



SUMMARY OF THE IMPORTANT POINTS OF THE 
CHAPTER 

It is important to have clearly in mind the following im- 
portant conclusions from the chapter : 



Finding Unknown Distances by Similar Triangles 95 

1. If you know that the angles of one triangle are 
equal respectively to the angles of another tri- 
angle, then you know that the ratios of the corre- 
sponding sides are equal In other words, you can 
make an equation, and thereby find an unknown 
side. 

2. If you know that the ratios of the corresponding 
sides of two triangles are equal, then you know 
that the angles of one triangle are equal respec- 
tively to the aggies of the other triangle. 

3. The corresponding angles of parallel lines are 
equal. 

4. Unknown distances may be found by means of 
similar triangles. 



REVIEW EXERCISE 39 

1. Translate into words : \y + 3 = y + 21. 

2. If A, B y and C represent the number of degrees 
in the respective angles of a triangle, we know 
that A + B + C = 180°. Why ? What is A if 
£ = 40 D and C=65°? 

3. If five times a certain number is divided by 2.7, 
the result is 3. What is the number ? 

4. Given the formula V= lwh> find a formula for 
/; for w. 

5. A boy receives C cents an hour for regular work, 
and pay for time and a half when he works over- 
time. What will represent his earnings for 
6 hr. overtime? Evaluate this when £7=50. 



96 Fundamentals of High School Mathematics 

6. The number of years that a man at various ages 
may expect to live, as determined by insurance 
experts, is as follows : 



If a man is 
(age in. years) 


10 


15 


20 


25 


30 


35 


40 


45 


50 


he may still 
live (years) 


49.6 


45.2 


41. 


37 


33.1 


29.2 


25 6 


22.2 


18.9 



Construct this graphically, representing ages on 
the horizontal axis. 

7. From the graph, find how much longer a man 
21 years old may expect to live ; a man 32 years 
old. 



CHAPTER VII 

HOW TO FIND UNKNOWNS BY MEANS OF THE RATIOS 
OF THE SIDES OF THE RIGHT TRIANGLE 

Section 38. The advantage of the RIGHT TRIANGLE in 
finding unknowns. It should be clear by this time that 
mathematics gives us methods of finding unknown quanti- 
ties. The equation is the most important tool for doing 
this, for the reason that when we solve a problem we have 
to make an equation. This equation must contain the un- 
known quantity together with other known quantities 
which are related to it in some way. 

In the last chapter we saw that an equation could be 
formed from the ratios of corresponding sides of similar 
triangles and that by that means we could find an un- 
known length. Two facts, however, make that method 
less satisfactory than the one we shall study in this 
chapter: (1) we must always be certain the triangles are 
similar, or we have no right to make an equation, and 
(2) the method is cumbersome because we must always 
use two triangles. 

There is a particular kind of triangle whose properties 
can be used to find unknown distances accurately and 
at the same time more easily than by any other method. 
It is the right triangle. The most important fact about 
the right triangle is found in connection with the ratios 
of its different sides. 

I. THE TANGENT OF AN ANGLE 

Section 39. The ratio of the " side opposite " a given 
angle to the " side adjacent " the given angle, i.e. the 
TANGENT of the angle. You will recall that in a right tri- 
angle one angle is 90° and the sum of the two acute angles 
equals 90°. (Why?) In finding unknown distances by 

97 



gS Fundamentals of High School Mathematics 

means of right triangles we shall always deal especially 
with one of the acute angles. Therefore, in referring to 




A side adjacent to 30 °L ^ 



Fig. 69 




° hypotenuse 

Fig. 70 

the sides of a right triangle, when dealing with a given 
angle, we shall speak of them as they are described in 
Figs. 69 and 70. If angle B is the acute angle with which 
we are concerned, then side AC is the " side opposite" 
Z B, and side AB is the " side adjacent " Z B. The side 
opposite the 90° angle is always called the hypotenuse. 
Some exercises will show the importance of the r&tio 

the "s ide o pposite " 
the "side adjacent" 

an acute angle of a right triangle. 



Finding Unknowns by Ratios of Sides of Triangle 99 

EXERCISE 40 

SOME EXPERIMENTS TO FIND THE NUMERICAL VALUE OF THE RATIO OF 
THE ■ ■ SIDE OPPOSITE ' » TO THE ' ■ SIDE ADJACENT " A 3CP ANGLE OF 
A RIGHT TRIANGLE 

T 




Fig. 71 



In Fig. 71, a t is the " side opposite " the 30° angle 
and b 1 is the "side adjacent" the 30° angle. 
Measure a x and b v Now find the numerical value 
of the ratio of a x to b x by dividing the length of 
a x by the length of b v Record your results in 
Table 3. 

In Fig. 72, a x and b x are respectively the " side 
opposite" and the "side adjacent" an acute 
angle of 30°. Measure each and compute the 

ratio ~r to 

b \ 
results in Table 3. 



two decimal places. Record your 




Fig. 72 



ioo Fundamentals of High School Mathematics 

3. Draw any other triangle similar to those above, 
but with much larger sides. Measure the "side 
opposite " and the " side adjacent" the 30° angle 
and compute their ratio as before. Record re- 
sults, as before, in Table 3. 

Table 3. Record here the results of measuring 
the sides of right triangles and of computing the 
ratio of the " side opposite " to the " side adja- 
cent " an acute angle 0/"3O°. 

Table 3 





Length of a a 


Length of b x 


Ratio of a^ob^e.,-^) 


Fig. 








% 








fig- 








Fig. 









What do you notice in the table about the nu- 
merical values of the ratios 

the "side opposite" an acute angle of 30° a 1? 

the "side adjacent" an acute angle of 30° b x 

The members of the class should compare re- 
sults, to see what result seems most likely to be 
the true one. If great care is taken in measur- 
ing, the ratio should be very close to .58 in each 
triangle. Why should it be the same in each 
triangle ? 



Finding Unknowns by Ratios of Sides of Triangle 101 



4. In Fig. 73, CB, DE, and GF are perpendicular to 
AB. Is tri- 



angle AFG 






V. 


similar to tri- 






D^^ 


angle AED1 




G^ 




Why? Is 




^*^ i 




either of the 




s^ i 




smaller tri- 


K s^ 


i 




angles sim- 




F 


E I 


ilar to the 




Fig. 73 





large triangle ? Why ? From this, why does the 

GF 

ratio of GF to AF, or — -, equal the ratio of DE 

AF 

DE 

to AE, or ? If you measured these lines. 

AE J ■ 

and computed the ratios, what would you expect 

to be true of the results ? 

Section 40. This last example is very important, because 

GF 

it shows, without measurement, that the ratio equals 

AF 



the ratio 



BE 
AE 



But this is the same as saying that the 



ratio of the "side opposite " to the "side adjacent" a 30° 
angle in one right triangle is ALWAYS equal to the ratio of 
the "side opposite" to the "side adjacent " a 30° angle in 
any other right triangle. The length of the sides may be 
far different, but the ratios should be the same. This 
shows that the ratios obtained in the table should have been 
the same, if it were possible to draw and measure without 
error. 

We shall now make use of the fact that the numerical 
value of the ratio of the "side opposite" to the "side ad- 



102 Fundamentals of High School Mathematics 

jaccnt" an acute angle of 30° (in a right triangle) is ap- 
proximately .58, 

EXERCISE 41 

l. A man wishes to determine the height of a 
smokestack. He finds that the angle of eleva- 
tion of the top of the smokestack, from a point 
200 ft. from the base of the smokestack, is 30°. 




Fig. 74 



Solution : 



= .58. 



200 
h = 200 x .58. 
h = 116 ft. 



(Why ?) 

(Why ?) 



Note here that — is the ratio of the " side opposite " to 

200 ** 

the "side adjacent" the 30° angle. From previous work 
we know that this ratio is .58. Thus, we can make the 

equation =.58. 

H 200 

2. In triangle ABC, angle A is 30° and angle C is 
60°. Find CB if AB is 75 yards. What is CB 
if AB is 10 inches ? Draw the figure. 

3. In triangle XYZ, angle X is 30° and angle Z 
is 60°. Find X Y if YZ is 116 ft. 



Finding Unknowns by Ratios of Sides of Triangle 103 

4. In Fig. 75, CD bisects angle C and is perpen- 
dicular to AB. How many degrees in angle 
BCD ? If CD is 100 cm., how long is DB ? 




In the right triangle ABC, angle B is 60°, angle 
A is 30°, and BC is 50 ft. Find AC 

In triangle XYZ, angle X is 30°. What do 

ZY 

you know about the ratio ^77^? State defi- 
nitely when this ratio is equal to .58. 




Fig. 76 

Section 41. It is convenient to name important ratios. 

Since it is helpful to use the ratios of the various sides of a 
right triangle, very frequently in finding unknown dis- 
tances, each is given a definite name. The ratio of the 
"side opposite " an acute angle to the "side adjacent" is 
called : 



io4 Fundamentals of High School Mathematics 



THE TANGENT OF THE ANGLE 

Its abbreviation is tan. Thus, in the above examples 
the tangent of an angle of 30° is constant; it is approxi- 
mately .58. 

EXERCISE 42 

1. Construct a right triangle such as Fig. 77, with 
AB equal to 4 cm. and angle A equal to 40°. 
Then measure BC and from that Q 
find the tangent of an angle of 
40°. Compare results with those 
of other members of the class. 

2. In a similar way 
find the tangent 
of an angle of 
50°. (Use AB 
as 4 cm.) Also 
find, the tangent 
of each of the fol- 
lowing angles : 60°, 70°, and 20°. 

Section 42. Summary of steps in finding the tangent of 
an angle. These examples show how to find the tangent of 
any angle. TJiree steps are necessary ; namely : (1) meas- 
ure the side opposite the particular angle ; (2) measure the 
side adjacent the angle; (3) divide the first number ob- 
tained by the second. To do this, however, for angles of 
all sizes from very small to very large, would require a 
great deal of labor, and probably give, for a great many of 
you, inaccurate results. To save this trouble, and at the 
same time get very accurate results, these ratios or tangents 
have been computed very carefully and compiled in a table 
like Table 4. (See Table of Tangents on page 105.) 




Fig. 77 



Finding Unknowns by Ratios of Sides of Triangle 105 



Table 4 
TABLE OF SINES, COSINES, AND TANGENTS 

Numerical Values of the Tangents, Cosines, and Sines of the 
Angles from 0° to 90° Inclusive 



Deg. 


tan 


cos 


sin 


Deg. 


tan 


cos 


sin 





.000 


1.000 


.000 


46 


1.04 


.695 


.719 


1 


.017 


.999 


.017 


47 


1.07 


.682 


.731 


2 


.035 


.999 


.035 


48 


1.11 


.669 


.743 


3 


.052 


.999 


.052 


49 


1.15 


.656 


.755 


4 


.070 


.998 


.070 


50 


1.19 


.643 


.766 


5 


.087 


.996 


.087 


















51 


1.23 


.629 


.777 


6 


.105 


.995 


.105 


52 


1.28 


.616 


.788 


7 


.123 


.993 


.122 


53 


1.33 


.602 


.799 


8 


.141 


.990 


.139 


54 


1.38 


.5S8 


.809 


9 


.158 


.988 


.156 


55 


1.43 


.574 


.819 


10 


.176 


.985 


.174 


















56 


1.48 


.559 


.829 


11 


.194 


.982 


.191 


57 


1.54 


.545 


.839 


12 


.213 


.978 


.208 


58 


1.60 


.530 


.848 


13 


.231 


.974 


.225 


59 


1.66 


.515 


.857 


14 


.249 


.970 


.242 


60 


1.73 


.500 


.866 


15 


.268 


.966 


.259 


















61 


1.80 


.485 


.S75 


16 


.287 


.961 


.276 


62 


1.88 


.469 


.883 


17 


.306 


.956 


.292 


63 


1.96 


.454 


.891 


18 


.325 


.951 


.309 


64 


2.05 


.438 


.899 


19 


.344 


.946 


.326 


65 


2.14 


.423 


.906 


20 


.364 


.940 


.342 


















66 


2.25 


.407 


.914 


21 


.384 


.934 


.358 


67 


2.36 


.391 


.921 


22 


.404 


.927 


.375 


68 


2.48 


.375 


.927 


23 


.424 


.921 


.391 


69 


2.61 


.35S 


.934 


24 


.445 


.914 


.407 


70 


2.75 


.342 


.940 


25 


.466 


.906 


.423 


















71 


2.90 


.326 


.946 


26 


.488 


.899 


.438 


72 


3.08 


.309 


.951 


27 


.510 


.891 


.454 


73 


3.27 


.292 


.956 


28 


.532 


.883 


.469 


74 


3.49 


.276 


.961 


29 


.554 


.875 


.4S5 


75 


3.73 


.259 


.966 


30 


.577 


.866 


.500 


















76 


4.01 


.242 


.970 


31 


.601 


.857 


.515 


77 


4.33 


]225 


.974 


32 


.625 


.848 


.530 


78 


4.70 


.208 


.978 


33 


.649 


.839 


.545 


79 


5.14 


.191 


.982 


34 


.675 


.829 


.559 


SO 


5.67 


.174 


.985 


35 


.700 


.819 


.574 


















81 


6.31 


.156 


.988 


36 


.727 


.809 


.588 


82 


7.12 


.139 


.990 


37 


.754 


.799 


.602 


83 


8.14 


.122 


.993 


38 


.781 


.788 


.616 


84 


9.51 


.105 


.995 


39 


.810 


.777 


.629 


85 


11.4 


.087 


.996 


40 


.839 


.766 


.643 


















86 


14.3 


.070 


.998 


41 


.869 


.755 


.656 


87 


19.1 


.052 


.999 


42 


.900 


.743 


.669 


88 


28.6 


.035 


.999 


43 


.933 


.731 


.682 


89 


57.3 


.017 


.999 


44 


.966 


.719 


.695 


90 


Inf. 


.000 


1.000 


45 


1.000 


.707 


.707 











io6 Fundamentals of High School Mathematics 



EXERCISE 43 
FINDING ANGLES AND TANGENTS FROM THE TABLE OF TANGENTS 

Find, from Tabic 4, each of the following: 

1. tan 4i>°. 

2. The angle whose tangent is .58. 

3. tan 57°. 

4. The angle whose tangent is .94. 

5. tan 14°. 

6. The angle whose tangent is f . 

7. tan 25°. 

8. The angle whose tangent is f . 

9. tan 45°. 

EXERCISE 44 
EXAMPLES WHICH INVOLVE THE USE OF THE TANGENT OF AN ANGLE 

1. Illustrative example. The brace wire AC of a telephone 
pole BC, Fig. 78, makes with 
the ground an angle of 62°. 
It enters the ground 15 ft. 
from the foot of the pole. 
Find the height of the pole BC. 
Solution : 
BC 
AB 
BC_ 
15 
BC= 15 x 1.88 = 28.2. 



= tangent 62°. 
= 1.88 (from the Table). 



2. 



The angle of elevation 
of the top of a tree, from 
a point 75 ft. from its 
base (on level ground), & -1 
is 48°. How high is the 
tree ? 




Fig. 78 



Finding Unknowns by Ratios of Sides of Triangle 107 



3. From a vertical height of 1500 yd. a balloonist 
notes that the angle of depression of the enemy 
trench is 51°. Find the distance from the trench 
to the point on the level ground directly below 
the balloonist. Make a drawing. 

4. The angle of elevation of an aeroplane at point A 
on level ground is 44°. The point B on the 
ground directly beneath the aeroplane is 450 yd. 
from A. How high is the aeroplane? 

5. If a flagpole 42 ft. high casts a shadow 63 ft. 
long, what is the angle of elevation of the sun ? 

6. In Fig. 79, CD is per- 
pendicular to AB. 
Find AD if angle 
^ = 60° and CD = 20. 

7. From the point of 
observation on a mer- 
chant vessel, the an- 
gle of depression of 
the periscope of a 

submarine was 17°. How far was the submarine 
from the merchant vessel, if the observer was 
40 ft. above the water ? 

8. Turn back to page 79 and solve problem 15 by 
this method. How do your results compare with 
those obtained by scale drawings ? 




10S Fundamentals of High School Mathematics 



II. THE COSINE OF AN ANGLE 



Section 43. The ratio of the "side adjacent " the given 
angle to the hypotenuse of the triangle, i.e. the COSINE. 

In the previous section we found that the ratio of the 
"side opposite" to the "side adjacent" an acute angle 
of a right triangle is always constant for any particular 
angle. This enabled us to find the length of the sides 
and the size of the acute angle. Now we come to another 
fact about right triangles. Let us examine a problem 
which cajinot be solved by the use of the tangent. 



In Fig. 80, BC repre- 
sents a telephone 
pole, AC an anchor 
wire, and ABthe dis- 
tance from the foot of 
the pole to the point 
at which the wire 
enters the ground, 
20 ft. The wire 
makes an angle of 
30° with the ground. 
How long is the wire? 







Fig. 80 



Clearly, AC cannot be found by means of the ratio which 
we called the tangent, because the tangent of 30° makes 
use only of BC and AB y and we must get a ratio which 
contains AC, Therefore, to solve this problem we shall 
have to learn how to use the ratio of tJie "side adjacent" 

AB 
the 30° angle, to tJie hypotenuse, or — - . 

Au 



Finding Unknowns by Ratios of Sides of Triangle 109 

EXERCISE 45 

EXPERIMENTS TO DETERMINE THE NUMERICAL VALUE OF THE RATIO 
THE "SIDE ADJACENT" A 30° ANGLE /£ THE C0SINB 



THE HYPOTENUSE OF THE TRIANGLE 




1. Measure the length of b x and c 1 in Fig. 81. 
Then compute the ratio — to two decimals. 

2. Draw any other triangle similar to Fig. 81, but 
with much longer sides. Find, as in Example 1, 
the ratio of the side adjacent the 30° angle, 
to the hypotenuse. Compare your result with 
that of Example 1. 




Fig. 82 



3. In Fig. 82, EF and GH are perpendicular to AB. 



Why does 



AH = AF = AB 
AG AE AC 



i to Fundamentals of High School Mathematics 

Section 44. The COSINE of a particular angle is CON- 
STANT. This shows that the ratio of the " side adjacent " 
a 30° angle to the hypotenuse of one right triangle is 
equal to the same ratio in any other right triangle which 
has an acute angle of 30°. For this reason, you would get 

the same numerical value for - 1 in Examples 1 and 2, if it 

c l 
were not for errors in measurement. 

Therefore, just as in the case of the tangent, so the 

" side adjacent " 30° ansrle . , 

cosine, i.e. '- - — , is always constant, 

hypotenuse 

when the angle is 30°. It is approximately .86. The 

right triangles may differ in size and position, but as long 

as they are similar (that is, so long as the acute angles we 

are dealing with are the same size), this ratio does not 

change. 

EXERCISE 46 

PROBLEMS SOLVED BY APPLYING THE CONCLUSION ARRIVED AT ABOVE; 
NAMELY, THE RATIO OF THE "SIDE ADJACENT" A 30° ANGLE TO 
THE HYPOTENUSE IS .86. 

1. Illustrative example. The anchor wire AC, of a 
telephone pole, meets the ground 20 ft. from the foot of 
the pole, making an angle of 30° with the ground. Find 
the length of the wire AC. 

Solution: ^=.86. (Why?) r 

20 



.86. 


^s' 


.86 h, or h = 23.2 ft. 


y^ 


kS* 


30° 

1 



_ B 

Fig. 83 



Finding Unknowns by Ratios of Sides of Triangle 1 1 1 

2. The rope, AC> of the flagpole, BC, makes an 
angle of 30° with the ground, at a point 42 ft. 
from the foot of the pole. How long is the 
rope ? Make a drawing. 

3. The angle of elevation of the top of a tree from 
a point A y on level ground, 100 ft. from the 
base of the tree, is 30°. What is the distance 
from A to the top of the tree ? 

4. In the right triangle ABC, AB is 64 cm. and 
ZA = 30°. Find AC 




-64 cm. 

Fig. 84 

5. Draw a right triangle such that angle A = 60° 
and the hypotenuse AC =60 cm. From this 
could you find BC? 

6. The angle of depression of a boat, from the top 
of a cliff, is 30°. Find the distance from the 
observer to the boat, if the boat is 400 ft. from 
the foot of the cliff. 

These examples have been solved by using the ratio 
of the "side adjacent" an acute angle of 30° to the 
hypotenuse, or, as we shall call it from now on, by using 
the cosine of the angle. The abbreviation for cosine is 
cos. Thus, 

, D ratio of " side adjacent "ZB 
cosZ B = —± — • 

hypotenuse of the triangle 



112 



Fundamentals of High School Mathematics 



EXERCISE 47 



l. Construct a right triangle similar to Fig. 85, with 
A = 40° and AB = 4 cm. Then measure c x and 



compute the ratio 



ill. 



By comparing your 



result with cos 40° as given in the table, see 
if you are within .05 of the correct result. 

C 




l< — 1^=4 cm. 

Fig. 85 



B 



2. 



How would you construct or draw the cos of 

a 60° angle ? of an 80° angle ? 

Read from the table of cosines : 

(a) cos 67°. 

(&) The angle whose cos is .258. 

(c) cos 45°'. 

(d) The angle whose cos is .573. 

(e) cos 2°. 

(/) The angle whose cos is .707. 
\g) cos 89°. 

(//) The angle whose cos is .629. 
A surveyor desires to measure the distance EC 
across a swamp. He surveys the line BA per- 
pendicular to BC. He extends this line BA until 
he can measure from A to C. If AC is 400 ft. 



Finding Unknowns by Ratios of Sides of Triangle 113 



7. 



and angle C is 55°, show how he would com- 
pute the length of BC. Find BC 




A boy observes that his kite has taken all 
the string, 750 ft. Assuming that the string is 
straight and that it makes an angle of 34° with 
the ground, how far on level ground is it from 
the boy to the point directly below the kite ? 
The angle of elevation of the top of a tent pole, 
from a point 43.2 ft. from the foot of the pole, 
is 32°. Find the distance from the point of ob- 
servation to the top of the pole. 
Figure 87 is a risfht triangle. 



angle A is 42° and AC = 61 
is the cosine of angle A ? 



Find AB if 
Hint : What 




Fig. 8' 



ii4 Fundamentals of High ScJiool Mathematics 



8. How long a rope will be required to reach from 
the top of a flagpole to a point 19 ft. from the 
foot of the pole (on level ground) if the rope 
makes an angle of 63° with the ground ? 

9. The aogle of depression of a boat from the top 
of a cliff is 37° when the boat is 1260 ft. from 
the foot of the cliff. Find the distance from 
the boat to the top of the cliff. 

10. Find angle A if AB is 27 and AC is 48. 
Hint : What is \\ with respect to angle A ? 



ll. 



12. 




Fig. 88 

A man starts at O and travels in a direction 
which is 48° east of a north-south line. How 
far due north of O will he be when he is 26 
miles from O ? 

From the table find the cosine of 32°. Then 
find the cosine of an angle twice as large as 32°, 
and see if it is twice as large as the cosine of 
'■\'l . Does the cosine of an angle change or 
vary in the same way that the angle changes 
or varies ? 



Finding Unknowns by Ratios of Sides of Triangle 115 

III. THE SINE OF AN ANGLE 

Section 45. The ratio of the " side opposite " an acute angle 
to the hypotenuse, or, the SINE of the angle. We have 
now used two particular ratios of the sides of a right 
triangle, the tangent and the cosine. By using them we 
were able to determine unknown lines and angles. But 
these two ratios are not sufficient to find any side or 




Fig. 89 



any angle of a right triangle. For example, we have no 
ratio which involves BC and AC in Fig. 89. This brings 
us to the third (and last) important ratio : 

the " side opposite" an acute angle 
the hypotenuse 

In the same way as before, we can show that the numeri- 
cal value of this ratio is constant for any given angle. 
Having discussed the tangent and cosine so completely, it 
is unnecessary to take the trouble to construct or to com- 
pute the value of this ratio. The numerical values of 
this ratio, for all acute angles, are given in the table of 

SINES. 



n6 Fundamentals of High School Mathematics 




FIG. 90 



EXERCISE 48 

EXAMPLES SOLVED BY MEANS OF THE SINE OF AN 
ACUTE ANGLE 

1. A man travels from O in a direction which is 50° 
east of a north-south 
line. How far is he 
from the north-south 
line when he has trav- 
eled 60 miles from 
the starting point, O ? 

2. How far was the man 
from an east-west 
line through the point 
O? 

3. To what height, on a 
vertical wall, will a 38-foot ladder reach, if it 
makes an angle of 58° with the ground ? 

4. An aviator, 4200 yd. directly above his own lines, 
takes the angle of depression of the enemy's bat- 
tery. What must be the 
range of the enemy machine 
guns to endanger him, if the 
angle of depression is 29° ? 

5. What ratio gives the sine 
of Z A in this figure ? co- 
sine C? If A is 60°, what 
is C? Compare sine 60° 
with cosine 80°, from the 
table. State in words your 
conclusion. FlG - 91 




Finding Unknowns by Ratios of Sides of Triangle 117 

REVIEW EXERCISE 49 

In this list of problems you will have to decide for your- 
self whether to use the tangent, the cosine, or the sine. 
Make a drawing for each problem ; indicate the parts that 
you know, and the part you are to find. 

1. A flagpole 50 ft. high casts a shadow 80 ft. 
long. What is the angle of elevation of the 
sun ? What time of year is it ? 

2. A searchlight on the top of a building is 180 ft. 
above the street level. Through how many 
degrees from the horizontal must its beam of 
light be depressed so that it may fall directly on 
an object 400 ft. down the street from the base 
of the building ? 

3. From the top of a cliff 120 ft. above the surface 
of the water, the angle of depression of a boat is 
20°. How far is it from the top of the cliff to 
the boat ? 

4. At a time when the sun was 55° above the hori- 
zon, the shadow of a certain building was found 
to be 98 ft. long. How high is the building ? 

5. A 40-foot ladder resting against a building makes 
an angle of 53° with the ground. Find the dis- 
tance from the foot of the ladder to the building, 
and the distance from the top of theJadder to the 
base of the building. 

6. A man starts at O and travels in a direction 
which is 24° west of a north-south line through 
O y at the rate of 80 miles per day. At the end 
of 4 days how far north is he from an east-west 



n8 Fundamentals of High School Mathematics 

line through O? How far west is he from a 
north-south line through O ? 

7. What direction will a boy be from his starting 
point if he goes 40 miles due north and then 
18 miles due east? 

8. The gradient or slope of the railroad which runs 
up Pike's Peak is, in some places, 18 %, i.e. in 
going 100 ft. horizontally it rises 18 ft. What 
angle does the road make with the horizontal ? 



CHAPTER VIII 

HOW TO SHOW THE WAY IN WHICH ONE VARYING 
QUANTITY DEPENDS UPON ANOTHER 

Section 46. Quantities that change together. We have 

already seen that there are many illustrations of quantities 
that change together. The amount of money paid out for 
rent at $30 per month changes with, or depends upon, the 
number of months ; the time required to walk a certain 
distance, say 10 miles, changes as, or depends upon, the 
number of miles one walks per hour. In other words, 
there are varying quantities which are so related that a 
change in the value of one of tliem causes a change in tlie 
value of the otJier. 

This chapter will deal with quantities which change 
together. In addition to what you already know about 
these varying quantities, we shall now study just how these 
quantities vary. For example, does an increase in the 
value of one varying quantity cause a corresponding in- 
crease in the related quantity ? Or does an increase in one 
varying quantity cause a corresponding decrease in the 
other? Can- these be expressed (1) graphically, or (2) by 
tables, or (3) by formulas? These are the points which 
will be studied in the chapter. 

Section 47. Variables and constants. In our study of 
time, rate, and distance problems we saw that the distance 
traveled by a train running at any given rate changes or 
varies as the time which it has been running cJianges or 
varies. If a train runs at the rate of 40 miles per hour, its 
movement is described by the equation 

d=4:0t. 

In this equation, d and / change as the train progresses 
along its journey. The value of d depends upon the value 

119 



V 

i2o\j Fundamentals of High School Mathematics 

of t. This means that the distance and time are variables, 
while the rate is constant. 

Table 5 shows the tabular method of representing the 
relation between these related variables. This shows that 



Table 5 



If the no. of hrs. is 


1 


2 


3 


4 


5 


8 


10 


IS 


20 


then the distance is 


40 


80 


120 160 200320 400 600 800 



a change in the time causes a change in the distance, or 
that a change in one variable causes a change in the related 
variable. 



EXERCISE 50 

1. In the above table, does an incirase in the num- 
ber of hours always cause an increase in the dis- 
tance ? 

2. In the same table, find the ratio of each distance 
to its corresponding time. How do these ratios 
compare ? Do the ratios change ? 

3. A man buys a railroad ticket at 3 cents per 
mile. Show by the tabular method the relation 
between the cost and the number of miles 
traveled. Show from the table that as the dis- 
tance increases the cost increases, but that the 
ratio of the cost to the distance does not change. 
What equation will show the same thing the table 
shows ? 

4. Write the equation for the cost of any number 
of pounds of sugar at 9 cents per pound. What 
are the variables in your equation ? Tabulate 



How One Varying Quantity Depends on Another 121 

the cost for 1, 2, 5, 8, and 10 pounds. Show 
from the table that the ratio of the cost to the 
number of pounds does not change ; that is, it is 
constant, 

5. A rectangle has a fixed base, 5 inches. Its alti- 
tude is subject to change. Tabulate its area if 
its altitude is 4, 6, 8, 10, and 12 inches. Com- 
pare the ratio of any two values of the area with 
the ratio of the two corresponding values of the 
altitude. If one altitude is three times another 

altitude, the one area is I times the other 

area. Write the equation for its area. 

6. A bicyclist rides 10 miles per hour. Show, by 
three methods, the relation between the number 
of miles he travels and the number of hours 

required. In 6 hours he travels • times as 

far as he travels in 8 hours. 

I. DIRECT VARIATION, OR DIRECT PROPORTIONALITY 

Section 48. The problems in the previous exercise illus- 
trate direct variation, or direct proportionality. In each of 
the examples, one of the variables depended upon another 
variable for its value, and the ratio of any tzvo values of 
one variable was equal to the ratio of the two corresponding 
values of the other variable. When two variables are 
related in this way, one is said to vary as or to be directly 
proportional to the other. Thus, to prove that two varia- 
bles are directly proportional, or vary directly, we must 
show that 

The ratio of any two values of one variable is equal to 
the ratio of the two corresponding values of the other 
variable. 



[22 Fundamentals of High School Mathematics 

EXERCISE 51 

1. Illustrative example. A man earns $ 6 per day. Show 
that the amount he earns is directly proportional to the num- 
ber of days he works. 

Solution : 

(1) A = 6 d. (We write the equation first, from the condi- 
tions of the problem.) 

(2) Tabulating : 

Table 6 



If dis 


l 


2 


5 


8 


10 


12 


then A is 


6 


12 


30 48 60 


72 



(3) Now select any two values of A, say 12 and 60, and the 
two corresponding values of d, which are 2 and 10. If the ratio 
of these two values of A is equal to the ratio of these two 
values of d, then in the equation A — 6 d we know that A 
is directly proportional to d, or that A varies directly as d. 
Does J5 = T %? Yes. 

Thus, A is directly proportional to d, or the amount a man earns 
at 1 6 per day is directly proportional to the number of days 

A x rfi 



he works. This is often written — = . 

A 2 d 2 



Ai means some 

particular value of A, and A 2 means some other particular 
value of A; d\ and d> mean those particular values of d 
which correspond to the selected values of A\ and^4 2 - 



2. Write the equation for the area of a rectangle 
whose base is 10 inches. Then show by select- 
ing particular values of A and k that the area 
is directly proportiojial to the altitude. In other 
words show that 



Hoiv One Varying Quantity Depends on Another 123 

3. Write the equation for the circumference, C, of 
a circle whose diameter is D. Is C directly pro- 
portional to D ? Why ? 

4. Show that the area of a square is directly pro- 
portional to the square of its side. 

5. Write the equation for the area of a circle. 
Show that the area varies directly as the square 
of the radius. 

6. Show that the interest on 11000 at 6% is di- 
rectly proportional to the time. 

7. x varies directly as r, and when ,r = 1<), j' = -. 
Find the value of x when y = 7. 

8. C varies directly as d % and when cf= 12, e= 4 V . 
What is d when c = 72 ? 

9. Is your grade in mathematics directly propor- 
tional to the amount of time you spend in pre- 
paring your lessons ? 

10. Is the cost of a pair of shoes directly propor- 
tional to the size ? 

II. INVERSE VARIATION 

Section 49. When quantities are inversely related to each 
other. In the previous exercise the varying quantities 
were so related in any particular problem that an increase 
in one variable caused a corresponding increase in the other 
variable. Some variables, however, are so related that an 
increase in one is accompanied by a corresponding decrease 
in the other. 

An example : An increase in the rate at which a train 
moves causes a decrease in the time required to travel a 
certain distance. If the train travels at the rate of 20 miles 



If the rate is 


10 


121 


15 


20 


25 


30 


33| 


40 


50 


then the time is 


10 


8 


6f 


5 


4 


^2 
°3 


3 


2* 


2 



r24 Fundamentals of High School Mathematics 

per hour, it will require 5 hours to cover 100 miles ; but if 
it increases its rate to 30 miles per hour, it will decrease the 
time so that only 3^ hours will be required to make the 
trip. 

Let us illustrate this fact more in detail by tabulating 
the relation between the rate and the time of a train which 
makes a trip of 100 miles. Note from the table- how a 
change in one variable, say the rate, is accompanied by a 
change in the other variable, the time. 

Table 7 



This shows that an increase in the rate is accompanied 
by a decrease in the time. If we select any two values of 
the rate, say 20 and 50, and the corresponding values of the 
time, 5 and 2, we see that the ratio of the two values of the 
rate §{} is not equal to the ratio of the corresponding values 
of the time |. Clearly, |g does not equal |, or, to use the 
more general form, 

r t 

-1 does not equal -1. 



These ratios would be equal, however, if we should invert 
one of them, e.g. 



20 
50 



r or - 
5 ; 



l_ 



The fact that the ratio of any two values of one of the va- 
riables is equal to the inverted ratio of the corresponding 



How One Varying Quantity Depends on Another 125 

values of the other variable leads us to say that one of them 
is inversely proportional X.o the other, or varies inversely as 
the other. 

This gives the following principle: 

One variable is inversely proportional to another when the 
ratio of any two values of one of them is equal to the IN- 
VERTED RATIO of the two corresponding values of the 
other. 

EXERCISE 52 

1. The area of a rectangle is 200 sq. in. Show 
that the base varies inversely as the altitude, or 

b-i a 
that t- = — * 
b 2 a x 

2. The number of men doing a piece of work varies 
inversely as the time. If 10 men can do a piece 
of work in 32 days, in how many days can 4 men 
do the same work ? 

3. The variable y varies inversely as x, and when 
x = 12, y = 4. Find x when y = 16. 

4. Write an equation to show that the altitude and 
base of a rectangle, whose area is fixed, are in- 
versely proportional. 

Section 50. Graphical method of representing inverse va- 
riation. Figure 92, on the following page, shows graphi- 
cally the relation between two numbers which are inversely 
proportional, or which vary inversely. It represents the 
base and altitude of a rectangle whose area is always con- 
stant, say 100 sq. ft. 



126 Fundamentals of High School Mathematics 



(N 


I 






























I 

i 

\ 






























to 

QO 


\ 

i 

\ 






























CM 

r i* 1 


i 

1 






























WCM 

go 




1 


























































US 




\ 


' 






























\ 


























c0 








■•^. 


^> 


k^._ 




































■«»•«■ 


— * — 


1 


>—. 












O 4- 8 12 16 20 24- 28 32 36 ^K) 44 48 52 

BASE 

Fig. 92. The line shows the relationship between two numbers 
which vary INVERSELY ; in this case the relationship between 
the altitude and base of a rectangle whose area is constant, say 
100 sq. ft. As the altitude INCREASES, the base DECREASES. 

To construct this graph, the following table was made : 



Table 8 



If base is 


2 


4 


5 


6 


8 


io 


12.5 


20 


then altitude is 


50 


25 


20 


16.6 


12.5 


10 


8 


5 



Note that as the base increases, the altitude decreases. 
How does the ^raph show this relation ? In what way 
does this graph differ from those you have previously 
dealt with ? 



How One Varying Quantity Depends on Another 127 
Show that the equation 






describes the relation between the base and altitude of any 
rectangle whose area is constant, say 100 sq. ft. 



EXERCISE 53 
GRAPHICAL REPRESENTATION OF INVERSE VARIATION 

1. The product of two variables, x and 7, is always 
200. Tabulate 10 pairs of values of these vari- 
ables, and from the table construct a graph show- 
ing the way in which the variables are related. 
Measure values of x along the horizontal axis. 

2. Some tourists decide to make a trip of 100 miles. 
Show graphically the relation between (1) the 
different rates at which they might travel, and 
(2) the time required at each rate. 



SUMMARY 



1. Two related variables or changing quantities are 
directly proportional, or vary directly, when a 
change in one is accompanied by a corresponding 
change in the other. 

To test for direct variation, it is necessary to see 
whether the ratio of any two particular values of 
one variable is equal to the ratio of the two cor- 
responding values of the other variable. 



[28 Fundamentals of High School Mathematics 

2. Two related variables or changing quantities are 
inversely proportional, or vary inversely, when an 
increase in one is accompanied by a corresponding 
decrease in the other. 

To test for inverse variation, it is necessary to see 
whether the ratio of any two values of one vari- 
able is equal to the inverted ratio of the two cor- 
responding values of the other variable. 

3. The graph of direct variation is a straight line, 
while the graph of inverse variation is a curve. 

REVIEW EXERCISE 54 

1. If 60 cu. in. of gold weighs 42 lb., how much 
will 35 cu. in. weigh ? 

2. If a section of a steel beam 10 yd. long weighs 
840 lb., how long is a piece of the same material 
which weighs 1250 lb. ? 

3. At 40 lb. pressure per square inch, a given pipe 
discharges 160 gal. per minute. How many 
gallons per minute would be discharged at 
65 lb. pressure? 

4. A steam shovel can handle 900 cu. yd. of earth 
in 7 hr. At the same rate how many cubic 
yards can be handled in 5 hr. ? 

5. A train traveling at the rate of 50 miles per 
hour covers a trip in 5 hours. How long would 
it take to cover the same distance if it traveled 
at the rate of 35 miles per hour ? 

6. If 50 men can build a boat in 20 days, how long 
would it take -30 men to build it ? 



How One Varying Quantity Depends on Another 129 

7. A wheel 28 in. in diameter makes 42 revolutions 
in going a given distance. How many revolu- 
tions would a 48-inch wheel make in going the 
same distance ? 

8. The volume, v, of a gas is inversely proportional 
to its pressure, /. Write an equation showing 
this fact. 

9. If the volume of a gas is 600 cubic centimeters 
(cc.) when the pressure is 60 grams per square 
centimeter, find the pressure when the volume 
is 150 cubic centimeters. 

10. When are two changing quantities or variables 
directly proportional ? When do they vary 
inversely ? 

11. How can you test for direct variation ? for in- 
verse variation ? Are x and y directly propor- 
tional in the equation x = 2y-\-5? 

12. If you know that 

7£_6 = 2£ + 24, 

then what is done to each side of the equation 
to get 7 b = 2b + 30 ? 

What is the next step in solving this equation ? 
Find the value of b. How do you check it ? 

13. How can you get rid of fractions in the equation 

1^+5 = 1^+29? 
Why is 24 not the most convenient multiplier? 

14. In order to save d dollars in n years, how much 
would your savings have to average per month? 



CHAPTER IX 

THE USE OF POSITIVE AND NEGATIVE NUMBERS 

Section 51. We need numbers to represent opposite quali- 
ties, or numbers of opposite nature. The examples in the 
following exercises will illustrate what is meant by opposite 
qualities, or numbers of opposite nature. We shall take four 
different kinds of illustrations : (1) opposite numbers on 
a temperature scale, (2) opposite numbers on a distance 
scale, (3) opposite numbers to represent financial situations 
(" having " and " owing "), (4) opposite numbers on a time 
scale, to represent " time before " a beginning point and 
"time after." 

FIRST ILLUSTRATION: OPPOSITE NUMBERS ON A 
TEMPERATURE SCALE 

EXERCISE 55 

1. The top of the mercury column of a thermometer 
stands at zero degrees (0°). During the next 
hour it rises 3°, and the next it rises 4°. What 
is the temperature at the end of the second hour ? 

2. The top of the mercury column stands at 0°. 
During the next hour it falls 3°, and in the next 
it/alls 4°. What is the reading at the end of 
the second hour ? 

3. If it starts at 0°, rises 3°, and then falls 4°, what 
is the reading ? 

4. If it starts at 0°, falls 3°, and then rises 4°, what 
is the reading ? 

These examples show that we must distinguish two 
kinds of temperature readings, (1) those above zero and 
('1) those below zero. People have agreed to call read- 
ings above zero " POSITIVE" and readings below zero 

130 



The Use of Positive and Negative Numbers 131 

"NEGATIVES Thus, if the mercury starts at zero and 
rises 4°, it will be at positive 4°, or, more briefly, + 4°. 
But if it starts at zero and falls 4°, it will be at negative 
4°, or — 4°. In the remainder of these examples you 
should describe the mercury readings as positive or nega- 
tive, rather than as above or below zero. 

5. The temperature stands at zero. Its first change 
is described by the expression + 6°. Its next 
change is described by + 4°. What is the tem- 
perature at the end of the second change ? 

6. If the temperature reading is 0°, and it makes 
the change — 5°, then — 3°, what is the final 
reading ? 

SECOND ILLUSTRATION: OPPOSITE NUMBERS ON A 
DISTANCE SCALE 

EXERCISE 56 

1. An autoist starts from a certain point and goes 
east 10 miles, and then east 8 miles. How far 
and in what direction is he from the starting 
point ? 

2. If he had first gone west 10 miles, and then west 
8 miles, how far and in what direction would he 
have been from his starting point ? 

3. If he had first gone east 10 miles and then west 
8 miles, how far and in what direction would he 
have been from his starting point ? 

4. If he had first gone west 10 miles, and then east 
8 miles, how far and in what direction would he 
have been from his starting point ? 



132 Fundamentals of High School Mathematics 

These examples show that we must distinguish between 
opposite distances, those cast of some starting point, and 
those west of the starting point. People have agreed to 
call distances cast of the starting point positive and distances 
west of the starting point negative. By this means a great 
deal of time can be saved, because a positive or negative 
number tells both the direction and the distance of a point on 
the distance scale, from some beginning point. Thus, on 
the distance scale, Fig. 93, point A is completely described 
by the number — 5. 

West A East 

1 1 \ T 1 1 1 1 1 — I 

-20 -15 -10 -5 +5 +10 +15 +20 +25 

Fig. 93. Points on a distance scale. 

This number, — 5, tells that the point A is 5 units west 
of, or to the left of, the starting point. 

5. What would be the position on this distance 
scale of a man who starts at the zero point, goes 
east 60 units, and then west 15 units ? 

6. Where would you be if you started at zero, went 
+ 8 units, and then —8 units ? 

7. A man starts at ; at the end of the first day he 
is at + 20, and at the end of the second day 
he is at — 10. What is the total distance he 
traveled ? What number will completely de- 
scribe his position at the end of the second day? 

8. How far is it from + 9 to — 6 ? What direction 
is it ? 

Section 52. Thus, positive and negative numbers are 
used to distinguish between opposite qualities. The fore- 
going examples show that we need a brief, economical way 



The Use of Positive and Negative Numbers 133 

to denote opposite qualities of numbers. This is done by 
positive and negative numbers, or, as we shall say from 
now on, by signed numbers. Thus, in referring to tem- 
perature readings, e.g. the " signed " number, + 10°, shows 
(1) how far and (2) in what direction the mercury stands 
from the zero point. In describing the location of a point 
on a distance scale, the " signed" number, — 6, tells how 
far and in what direction the point is from the starting or 
zero point ; that is, 6 units to the left of, or to the west of, 
the zero point. 

THIRD ILLUSTRATION: OPPOSITE NUMBERS TO 
REPRESENT FINANCIAL SITUATIONS 

Section 53. Positive and negative numbers, or SIGNED 
NUMBERS, are used also to describe financial situations. 
It has been agreed to consider money that you " have " as 
positive and money that you " owe " as negative. Thus, 
if you owe 40 cents (i.e. — 40 cents) and have 55 cents 
(+55 cents), your real financial situation is +15 cents. 
Why ? Or, if you owe 90 cents ( — 90 cents) and have 75 
cents (+ 75 cents), your ;r#/financial situation is — 15 cents. 

FOURTH ILLUSTRATION: OPPOSITE NUMBERS ON 
A TIME SCALE 

Signed numbers are used also to distinguish " time be- 
fore " from "time after" a given time. For example, if 
time before Christ is negative, then time after Christ is 
positive. Thus, on the time scale below, since Christ's 
birth is regarded as zero, if a man was born 10 years be- 

EC- , . M^fffl-T^ . AP+ 

25 -20 -75 -10 S O +5 +10 +15 +20 +25 +30 +35 
Fig. 94. Points on a time scale. 



134 Fundamentals of High School Mathematics 

fore Christ and lived 35 years, the distance between the 
points A and B would represent the period of his life. 
Why ? 

OTHER ILLUSTRATIONS OF THE USE OF SIGNED 
NUMBERS : FOR THE .PUPIL TO DEVELOP 

EXERCISE 57 

1. Show how signed numbers are helpful in deal- 
ing with latitude; with longitude. Illustrate 
each one. 

2. Show that signed numbers are a convenience 
in keeping scores in games in which you either 
make or lose a certain number of points. 

3. Can you think of any other illustrations of 
opposite numbers ? 

EXERCISE 58 
PRACTICE IN USING SIGNED NUMBERS 

1. Your teacher's financial situation is —$250. 
What does this mean ? 

2. A man's property is worth 15200 and his debts 
amount to $3300. How can positive and 
negative numbers be used to represent these 
amounts ? What number will describe his net 
financial situation ? 

3. The mercury at 8 a.m. was at — 6°. If it was 
rising 3° per hour, where was it at 9 a.m. ? at 
10 a.m. ? at 11.30 A.M.? 

4. Show on a time scale that Caesar began to rule 
the Roman people 31 years B.C., and ruled for 

I "> years. 



The Use of Positive and Negative Numbers 135 

5. What is the total number of miles traveled by a 
man who starts at zero on the distance scale if 
he is at +6 at the end of the first day, — 2 at 
the end of the second day, and at — 8 at the 
end of the third day ? 

6. On the distance scale, where would you be if 
you started at — 4 and went east 6 miles ? 

7. If your financial condition is + 60 cents, — 15 
cents, and — 12 cents, what single number will 
accurately describe your net financial situation ? 

8. What was your final score in a game in which 
you made the following single scores : + 15, 
- 8, - 10, + 14, and + 15 ? 

9. Represent on a distance scale (horizontal) the 
point where a man would be at the end of the 
third day if he started at zero and walked + 6 
miles on Monday, — 10 on Tuesday, and — 3 
on Wednesday. 

10. Find the net financial situation of a man who is 
worth the following: (a) + $5+f8 + $10 
-$6; (b) + 6d-10d-8d+lod. 

Section 54. Absolute value of positive and negative num- 
bers. The numerical value of a positive or negative num- 
ber, without regard to its sign, is its absolute value. For 
example, the absolute value of + 6 is 6 ; of + 17 is IT ; of 
— 9 is 9, etc. 

Section 55. We need to be able to add, subtract, multiply, 
or divide signed numbers. Now that we see clearly the 
practical ways in which positive and negative numbers are 
used we need to be able to solve problems which contain 



136 Fundamentals of High School Mathematics 

either kind. In all the examples which we have worked 
previously, only positive numbers have been used. Next, 
therefore, we must learn (1) how to combine signed num- 
bers {i.e. add them) ; (2) how to multiply them ; (3) how 
to find the difference between two signed numbers; and 
(4) how to divide signed numbers. We will take them up 
in that order. 



I. HOW TO COMBINE SIGNED NUMBERS — FINDING 
ALGEBRAIC SUMS 

Section 56. When the numbers are arranged vertically. 

In the example : " Find the net financial situation of a man 
who is worth the following: +|5, +$8, -$10, -$6," 
we found one signed number which described the man's 
net financial situation; namely, — $3. That is, we found 
one signed number which was the result of putting several 
signed numbers together. This process is called combining 
signed numbers, or finding the algebraic sum. Thus, to 
combine +4, — 2, — 6, and + 3, we must find one signed 
number which is the result of putting all of these together. 
Evidently, this must be — 1. Similarly, combining, or 
finding the algebraic sum of + 5 d and — 11 d f we get 
-6d. 

In each of the following examples, find the algebraic 
sum, i.e. find one number which will describe the result of 
putting all the separate numbers together. 



l. 










EXERCISE 59 




+ 16 


2. 


- Id 3. + 4x 


4. 5 a 


- $3 




+ 8d -Zx 


4a 


+ $4 




-\d - X 


— 6a 



The Use of Positive and Negative Numbers 137 



5. 


6jj/ 


6. 


+ 8 


7. 


- 10 


8. 


+ 3£ 




-by 




-5 




+ 13 




-5b 




- y 




-6 
+ 7 




- 8 
+ 5 




-6b 
+ 2b 


9. 


4a 


10. 


-Zx 


11. 


Wad 


12. 


— 7 xy 




— 6a 




— 5x 
+ Sx 




12 ab 




+ 11 xy 

- 2xy 


13. 


+ 21* 


14. 


- 4 


15. 


X 


16. 


3# 




-7 x 




+ 3 




2x 




--la 




+ 2 x 




+ 11 

- 8 
+ 1 




-6x 




la 
- 9a 



Section 57. When the numbers are arranged horizon- 
tally. The numbers to be combined are almost always 
written in a horizontal line, rather than in a vertical column. 
Combining these terms is done in the same way as if 
they were written in a vertical column. For example, 

17. 6 - 8 - 5 + 11 = ? 

18. -8-9 + 11 + 6 

19. - Qd+ 5d+9d- 2d 

20. 5 abc + 6 abc — 7 abc 

21. 4/- 3/+ 6/- 9/ 

22. + 8 t - 9 / - 6 1 + 12 1 

23. How have you found the algebraic sum of 
these numbers ? 



138 Fundamentals of High School Mathematics 

Section 58. Terms are either LIKE or UNLIKE : How 

to distinguish TERMS. Any algebraic expression, such as 
ax + b or x 1 + *&xy + 5, is made up of one or more num- 
bers separated from each other by + signs or by — signs. 
These numbers thus separated from each other are called 
terms. Thus, in ax + b there are two terms ; ax is one, 
b is the other ; while in x 2 + 2 xy — 5 there are three terms, 
x 2 , 2xy f and 5. Note carefully that a "term" includes 
everything between + or — - signs. 

In many algebraic expressions these terms are all like 
terms, and, as we learned in the previous section, can 
be combined or put together into one number or term 
which we called their algebraic sum. Thus, 4 d, + 5 d, 

— 8d, + 3 d are similar or like terms, and their algebraic 
sum is + 4 d. It is important to understand that because 
each letter in the expression represents the same thing 
these are like terms. In many cases, however, the terms 
of an algebraic expression are not all like terms. 

For example, consider : 4 boys, + 5 girls, + 8 boys, 

— 2 girls, or, using the initial letters of the words, 4 b + 
bg + 8 b — 2g. Evidently, these are unlike terms and 
cannot be combined into one number. However, the like 
terms in the expression can be combined ; that is, the 4- 4 b 
and +8^, giving 12 b, and the + bg and — 2g, giving 
+ Sg. Thus the expression 4 b + bg + 8 b — 2g can 
be simplified or expressed more briefly by combining like 
terms, giving 12 b 4- 3^-. From this illustration we see 
that the like terms of any algebraic expression can be 
combined, giving a simpler, briefer expression than the 
original one. 



The Use of Positive and Negative Numbers 139 

EXERCISE 60 

FURTHER PRACTICE IN COMBINING SIGNED NUMBERS: NUMBERS HAVING 
LIKE OR UNLIKE TERMS 

Write in the simplest or briefest form each of the 
following expressions : 

1. 2a + 3a — 6a + ±a 

2. 5 ft. + 6 in. - 2 ft. - 4 in. 

3. 7 yr. + 3 mo. — 2 yr. — 1 mo. 

4. ±b + 5c - 8b ~2c + b 

5. 6 a 2 + 3 a 2 - 7 a 2 + 4 a 2 

6. - 2x*- 5x*- 8x* + 2*3 

7. ax + 5 + 4 ax + 3 

8. 3 xy + 5 ab — 7 ,17 — 11 <z£ 

9. 2 £ 3 - 7 £ 3 + 5 js - #> 

10. 2r+8r + 3r- 10 r 

11. 4 * 2j + 5 ^ _ 8 ^ _ 3 a * b 

12. _6 + 4-8+6-9 + 2 

13. 5^+3-8^-4 + 4^ + 1 

14. a 2 b + 4 *2j _ 6 a 2 £ 

15. .27 + 3 - 8 *y - 9 + 2 4:7 + 7 

16. / + 2^-8/ + 4^ + 6/ + 5^ 



SUMMARY OF IMPORTANT PRINCIPLES CONCERNING THE 
COMBINING OF SIGNED NUMBERS 

Section 59. You have now worked many examples in 
finding algebraic sums. From your experience with such 
examples, complete these three sentences which tell how to 
combine signed numbers : 



140 Fundamentals of High School Mathematics 

1. To find the algebraic sum of two positive numbers, 

l the absolute values of the numbers, and 

give to the result a I sign. 

2. To find the algebraic sum of two negative numbers, 

! the absolute values of the numbers, and 

give to the result a 1 sign. 

3. To find the algebraic sum of two numbers having 

unlike signs, find the I of their absolute 

values, and give to the result the sign of the 
number having the I absolute value. 

II. HOW TO MULTIPLY SIGNED NUMBERS 

Section 60. The four ways to multiply signed numbers 
In arithmetic it was found that multiplication shortened 
the work of addition. For example, in adding 3 + 3 + 3 + 
3 + 3 + 3 + 3, the result is found most easily by multiply- 
ing 3 by 7, because 3 is taken 7 times. So, in algebra, it 
is equally desirable to multiply one signed number by 
another. 

There are four different ways in which we may have to 
multiply signed numbers. These are : 

(1) plus times plus, as in the example +4 times 
+ 2 == ? 

(2) plus times minus, as in the example + 4 times 

— 2 = ? 

(3) minus times plus, as in the example —4 times 
+ 2 = ? 

(4) minus times minus, as in the example — 4 times 

— 2 = ? 

By considering the following problems we can tell what 
meaning must be given to the multiplication of signed 
numbers. 



The Use of Positive and Negative Numbers 141 



A. ILLUSTRATIVE QUESTIONS BASED UPON THE SAVING 
AND WASTING OF MONEY 

EXERCISE 61 

1. If you save $5 a month ( + $5), how much better 
off will you be 6 months from now (+ 6) ? 
Evidently you will be $30 better off ( + $30). 
Thus, + 5 times + 6 = + 30. 

2. If you have been saving $5 a month ( + $5), how 
much better off were you 6 months ago (— 6)? 
Evidently you were $30 worse off (- $30) than 
you are now. Thus, + 5 times — 6 = — 30. 

3. If you are wasting $5 a month (— $5), how much 
better off will you be in 6 months from now 
(+6)? 

Evidently you will be $30 worse off ( — $30). 
Thus, — 5 times + 6 = — 30. 

4. If you have been wasting $5 a month ( — $5), 
how much better off were you 6 months ago 
(-6)? 

Evidently you were $30 better off (+$30). 
Thus, — 5 times — 6 = + 30. 

Summarizing : These problems based upon saving and 
wasting money have led to the following illustrative state- 
ments : 

1. + 5 times + 6 = + 30! 

2. + 5 times - 6 = - 30. 
3.-5 times + 6 = - 30. 
4.-5 times - 6 = + 30. 



142 Fundamentals of High School Mathematics 

B. ILLUSTRATIVE QUESTIONS BASED UPON 
THERMOMETER READINGS 

EXERCISE 62 

1. If the mercury is now at zero and is rising 2° 
per hour (+ 2), where will it be 4 hours from 
now (+4)? 

Evidently it will be 8° above zero (+ 8). Thus, 
+ 2 times + 4 = + 8. 

2. If the mercury Jias been rising 2° per hour ( + 2°) 
and is now at zero, where was it 4 hours ago 
(-4)? 

Evidently it was 8° below zero ( — 8°). Thus, 
+ 2 times — 4 = - 8. 

3. If the mercury is now at zero and is falling 2° 
per hour ( — 2°), where will it be 4 hours from 
now (+4)? 

Evidently it will be 8° below zero (- 8°). Thus, 

- 2 times + 4 = - 8. 

4. If the mercury is now at zero and has beenfall- 
i?ig 2° per hour (— 2°), where zvas it 4 hours ago 
(-4)? 

Evidently it was 8° above zero ( + 8°). Thus, 

- 2 times —4 = 4-8. 

Summarizing : these problems based upon the ther- 
mometer have led to the following illustrative statements : 

1. + 2 times + 4 = + 8. 

2. + 2 times — 4 = — 8. 
3.-2 times + 4 = - 8. 
4.-2 times — 4 = + 8. 



The Use of Positive and Negative Numbers 143 

A careful study of these illustrations will enable you to 
complete the following statements concerning multiplica- 
tion of signed numbers : 

1. A positive number multiplied by a positive number gives as 
a product a ? number. 

2. A positive number multiplied by a negative number gives as 
a product a ? number. 

3. A negative number multiplied by a positive number gives 
as a product a ? number. 

4. A negative number multiplied by a negative number gives 
as a product a 1_ number. 

5. The product of two numbers having like signs is ? 

6. The product of two numbers having unlike signs is ? 



II a. PARENTHESES ARE USED TO INDICATE 
MULTIPLICATION 

Section 61. Multiplication of two or more numbers is 
often indicated by placing the numbers within parentheses. 
Thus, " + 4 times -35" is often written " ( + 4)(- 35)." 
It is important to note that no sign or symbol is placed 
between the parentheses when multiplication is indicated. 







EXERCISE 63 






PRACTICE IN 


MULTIPLYING SIGNED NUMBERS 


1. 


( + 3)(+5) 


8. < 


r +5)(+4)(-2) 


2. 


(+6)'(-2) 


9. ( 


_3)(-6)(+2) 


3. 


(+10)(-2|) 


10. ( 


;-4)(-10)(-3; 


4. 


( + 6)(-9) 


11. ( 


:+§)(- 1) 


5. 


(-2)(-5) 


12. ( 


;-fx+i) 


6. 


(+8)(-|) 


13. ( 


;-7)(-6)(+2) 


7. 


(+12)(+6) 


14. ( 


;-2)(-2)(-2) 



144 Fundamentals of High School Mathematics 





15. (- 


3)( 


-3)(- 


-3) 






16. (+l)( + l)( + l)+(+l) 


17. ( 


-2)(-2) 






31. 


(- §)(+!) 


18. ( 


-■2)(-2)(-i 


0(- 


•2) 


32. 


(-■W-tt) 


19. ( 


3)(4)(5)(2) 






33. 


(_2)(-2)( + 2) 


20. ( 


+ 2)( + #5) 






34. 


( + *)(-tt) 


21. ( 


+ 3)(+7Z>) 






35. 


(-JX- io) 


22. ( 


; + (!)( + 5 ft.) 






36. 


-5-8 


23. ( 


-8)( + 6j) 






37. 


-7 -21 


24. ( 


+ f)(18«) 






38. 


(-6X-*) 


25. 


(-12X+J) 






39. 


2 -2 -(-3) 


26. ( 


IX- 18) 






40. 


(_1)(-1)(-1)(_1) 


27. ( 


.27)(-!) 






41. 


(2)(-3)( + 4)(-5) 


28. ( 


: + !)(- 25) 






42. 


(6)(|)(-1) 


29. ( 


:-32)(-|) 






43. 


(10)(5)(-^) 


30. ( 


:-iK+2±j) 






44. 


(IX- 1) 



II J. HOW TO USE EXPONENTS IN MULTIPLICATION 

Section 62. Suppose we had to find the product of 3x 2 
and 5.r 4 . It is important to keep in mind the meaning of 
exponents. 3 x 2 means 3 - x • x and 5 x 4 means 5 • x • x • x • x. 
Hence, Zx 2 times 5.r 4 or (Sx 2 )(5x A ) means 3 • x • x • 5 • x • x 
• x • .r, or 15 ;tr 6 . By the same reasoning, the product of 
+ 6 x 6 and - 7 x 4 is - 42 ^ 9 . 



The Use of Positive and Negative Numbers 145 

EXERCISE 64 

PRACTICE IN USING EXPONENTS IN MULTIPLYING SIGNED NUMBERS 

(ORAL) 

1. (5a 2 )(6a 4 ) 13. - 2y ■ 3 y 2 

2. (+7£)(-9^) 14. (_5J)(-2J») 

3. (+8)(2y>) 15. (-6^ 2 )(-7^) 

4. (6^)(2« 2 ) 16. (—8 *»)(£*) 

5. ( + Sabc)(oab) 17. ( + 10ytr)( - 2y&) 

6. 4 „r 3 . 5 x 1 - 2 ;r 5 is. 16 ;z 2 £ - {\ ab») 

7. j 4 • 5/ 3 19. a - b - b - a - a 

8. J^. 10. r 7 20. .r 4 ■ 2* 

9. 2aP.§ab 21. j • 5j/ 3 

10. a 2 b-ab 2 -aW 22. ^.fe 

11. ;tr • 2;r 3 23. 5 . x 1 

12. - 6 a • 3 a 24. 10 j 2 - T \y . y* 



III. HOW TO FIND THE DIFFERENCE BETWEEN SIGNED 
NUMBERS: SUBTRACTION 

Section 63. How " differences " are found in practical 
work. Clerks in stores have a method of making change 
or of finding the difference between two numbers that is 
very helpful in finding the difference between two signed 
numbers. For example, if a customer gives the clerk 50 
cents in payment for a 27-cent purchase, the clerk begins 
at 27 and counts out enough money to make 50 cents. If 
we use the same terms as were used in arithmetic, — 
namely, the subtrahend, minuend, and difference, — then 
we say, " The clerk begins at the subtrahend, 27 cents, 
and counts to the*minuend, 50 cents." 



1 46 Fundamentals of High School Mathematics 



First illustrative example. To illustrate this method of 
finding the difference between two signed numbers, let us 
consider this problem : 



On a certain day the mercury 
stands at — 4° in Chicago and 
at -f 13" in St. Louis. How 
much warmer is it in St. Louis, 
or what is the difference between 
-f 13° and - 4° ? Naturally, we 
do the same thing the clerk does, 
begin at the subtrahend and count to 
the minuend, i.e. we count from 
- 4° to + 13°, giving us + 17°. 
The difference is called positive 
because we counted upward. If 
we counted downward, the dif- 
ference would be called negative. 
This example is written as fol- 
lows : 

+ 13° minuend 
— 4° subtrahend 
+ 17° difference 



-4- 



+13 



Chicago 



St. Louis 



Fig. 95 a 



Second illustrative example. Subtract 
-I- 10 from — 5 by referring to the number scale. 
This means to find the distance from the sub- 
trahend to the minuend or from +10 to — 5. 
The distance from 10 above to 5 below is clearly 
15 ; and since the direction is downward, the 
difference is — 15. This example is written : 

— 5 minuend 
-f 10 subtrahend 



— 15 difference 



i The Use of Positive and Negative Numbers 147 

These illustrations are given merely to show that the 
difference between two signed numbers can always be 
found by counting on a number scale from the subtrahend 
to the minuend. The difference will be positive or nega- 
tive, depending upon whether the direction of counting is 
upward or downward. 

EXERCISE 65 

PRACTICE IN FINDING THE DIFFERENCE BETWEEN TWO SIGNED 
NUMBERS: SUBTRACTION 



1. +6 


-4 +8 +3 


+ 5 


-7 + 


13 + 9 


- 2 


+ 5 + 2 +10 


-8 


+ 1 + 


4 +14 


2. + 14 


-6° + 7 d 


+ 10 ft. 


-4 in. 


+ Sx 


-18 


+ 9 5 -2d 

+ ±b -he 


- 6 ft. 

10 jr 2 


— 7 in. 


-10* 


3. - Za 


— -3 x^y 


+ 10 adc 


-11a 


-2b + 6e 


-2.T 2 


+ 5x s y 
3x 


— \abc 


4. -2fi 


+ 5xf 


5a 


-2 be 


+ 2,fi 


+ 11 xf 


-2a 


X 


be 



5. From — 7 a take + 5 a. 7. From x take 1 x. 

6. Take - 13 b 2 from + 2 b 2 . 8. Take - be from 2 be. 

IV. HOW TO DIVIDE SIGNED NUMBERS 

Section 64. Division is the opposite of multiplication. 

You will have little or no difficulty in the division of 
signed numbers if you understand that division is just 
the opposite of multiplication. For example, if 4 x 2 = 8, 
then |=4. In this case 8 is the dividend, 2 is the divisor, 
and 4 is the quotient. In signed numbers, as well as in 



148 Fundamentals of High School Mathematics 

arithmetic, the dividend equals the quotient times the 
divisor. 

+ 8-*--2 = -4;or -"±-| = -4, because (- 2)(- 4) = + 8. 
- 8 -f- - 2 = + 4 ; or -^| = + 4, because ( - 2)(+ 4)= - 8. 
+ 8- + 2 = +4;or:-^ = + 4, because ( + 2)( + 4)= + 8. 

_8-s- + 2 = -4; or r-| = -4, because (+ 2)(-4)= -8. 

4- 2 



EXERCISE 66 
PRACTICE IN FINDING THE QUOTIENTS OF SIGNED NUMBERS 

Find the quotient in each of the following : 

4-10. +18 . -16 . -30 . -14 . +6 . -8 . +16 , 
' +2' -2' +4' -10' +7' +2' -1' -2' 

+ 15rf —18 ft. +16 mo.. + 25.r -10<z 



2. 



3' +3' -8 ' -5 ' +2 



3. (21-)+(-8); (_36)-*-(-9); (- 54)^+27); 
(-96)-K+12); (_21)4-<-7); (-60)^( + 12). 

, 10.r3 2iy 18 tf> 12 r 4 16/ 3 u 

4 - IT 5 3/ ; 6* 5 ^ ; ifp' Howcan y° u 

prove each of these ? 

+ 20/ . -27**. -34£¥= , -50^y . -72/z^ 
' -4j ' -9x*' -17£ 2 ' + 25jj/' -6j/w 5 ' 



The Use of Positive mid Negative Numbers 149 

EXERCISE 67 
COMPLETING STATEMENTS ABOUT DIVISION 

1. A positive number divided by a positive number 
gives as a quotient a 1 number. 

2. A negative number divided by a positive number 
gives as a quotient a 1 number. 

3. The quotient of two numbers having like signs 
is L_. 

4. The quotient of two numbers having unlike 
signs is • 

EXERCISE 68 

A REVIEW OF ADDITION, MULTIPLICATION, SUBTRACTION, AND 
DIVISION OF SIGNED NUMBERS 

This is a very difficult but a very important exercise. 

1. From the sum of 2 a and — 5 a take the differ- 
ence between — 3 a and + 8 a. 

2. Add the product of — 3 and + 5 to the quotient 
of - 18 and - 2. 

3. Take the sum of — 7 b and + 4 b from the dif- 
ference between — 6 b and + 11 b. 

4. To the quotient of — 21 and + 3 add the 
product of — 6 and + 7. 

5. From the sum of 7 / and — 10 / take the differ- 
ence between — 4 i and + 11 A 

6. Add the product of — 6 and + 9 to the quotient 
of - 28 and - 4. 

7. Take the sum of — 9 x and + Sx from the dif- 
ference between — 5x and + 13;tr. 



150 Fundamentals of High School Mathematics 

8. To the product of — 7 and +11 add the quo- 
tient of — 33 and + 3. 

9. From the sum of 8<r and — 14 <; take the differ- 
ence between — be and + 16*:. 

10. Add the product of — 12 and + 5 to the quo- 
tient of - 32 and - 4. 

11. Take the sum of — 8j and 3j/ from the differ- 
ence between — 7 y and + 12j/. 

12. To the product of — 8 and + 9 add the quotient 
of - 36 and + 6. 

13. From the sum of 12 b and — 16 b take the dif- 
ference between —lb and + 8 b. 

14. Add the product of — 8 and + 9 to the quotient 
of — 40 and — 5. 

15. Take the sum of — 11 1 and 7 / from the differ- 
ence between — 9 / and + 10 t. 

16. To the product of — 6 and + 13 add the quo- 
tient of - 42 and + 7. 

SUMMARY 

This chapter shows the need of signed numbers. It 
teaches how to : 

1. Combine or add signed numbers. 

2. Find the difference between two signed numbers; 
that is, to subtract one signed number from 
another. 

3. Find the product of signed numbers. 

4. Divide one signed number by another. 



The Use of Positive and Negative Numbers 151 

REVIEW EXERCISE 69 

1. The formula h = 25 + 1 (G — 4) is used to deter- 
mine the proper height of the chalk trough in a 
schoolroom. If h stands for the height in 
inches, and g stands for the number of the 
grade, find the height for Grade VIII; that is, 
when g= 8. What is the proper height for a 
third-grade room ? 

2. Evaluate the expression ab 2 + a 2 b if a = 2 and 
b = -3. 

3. Show that the sum of any two numbers having 
unlike signs, but the same absolute value, is zero. 
Give some illustrations. 

4. In a class of 25 pupils, 2 were conditioned and 
6 failed. Express the ratio of the number of 
pupils that succeeded to the total humber in the 
class. What percentage is this ? 

5. The ratio of y + 1 to 9 is equal to the ratio of 
y + 5 to 15. Find y. 

6. The number of posts required for a fence is 84 
when they are placed 18 feet apart. How many 
would be needed if they were placed 12 feet apart ? 

7. If I am now x years old, what does the follow- 
ing expression tell about my age : 2 x + 5 = 55 ? 



CHAPTER X 

THE COMPLETE SOLUTION OF THE SIMPLE EQUATION 

Section 65. What we have already learned about the 
equation. Since the equation is the most important opera- 
tion in mathematics, we must be able to solve quickly and 
accurately equations of any kind. Thus far we have 
learned two very important facts about equations : 

1. That if we do anything to one side of an equation, 
we must do the same thing to the other side. 

2. That an equation is SOLVED when a value of the un- 
known is found which satisfies the equation ; that is, one 
which makes the numerical value of one side equal to the 
numerical value of the other side. 

Furthermore, we have learned: (1) how to solve simple 
equations of the type 

6 b + 3 = 45, 
or, c + 5 c = 20 + c , etc. ; 

(2) how to get rid of fractions in an equation, e.g. of the 
type l^r+l^r — 1 = 3; 

(3) how to solve word problems, first by translating them 
into equations and second by solving the equations. These 
methods, which you have now mastered, are important first 
steps in the more important problem of learning how to 
solve equations of any kind. That is your task in studying 
this chapter. 

I. SOLVING KQUATIONS WHICH CONTAIN NEGATIVE 

NUMBERS 



Section 66. There are just two more steps that we must 
learn in using equations. First, we must be able to solve 
equations which contain negative numbers ; second, we 
must be able to solve equations which contain parentheses. 

152 



Complete Solution of the Simple Equation 153 

Negative numbers occur very commonly in equations. 
The following examples illustrate this fact. 

EXERCISE 70 

Write as equations, and solve each of the following 
examples : 

1. What number multiplied by 7 equals — 28? 

2. What number multiplied by —5 equals 20? 

3. If a certain number be added to 13, the result 
is 8. Find the number. 

4. A certain number increased by 10 equals — 5. 
Find the number. 

5. If 7 be subtracted from a certain number, the 
result is — 3. What is the number ? 

6. If negative four times a certain number gives 
22, what is the number ? 

These examples show how negative numbers occur in 
equations. Throughout the remainder of the work, equa- 
tions which are satisfied by negative numbers will occur 
very commonly. The next exercise contains many exam- 
ples of this kind. 

EXERCISE 71 

SOLUTION OF EASY EQUATIONS WHICH CONTAIN NEGATIVE NUMBERS 

Solve each of the following equations. You should be 
able to tell exactly what must be done to each side of the 
equation. 

1. ;r+5 = 3 5. 2 x+lQ = 2 

2. 2j=-16 6. -4^ = 12 

3. £ + 7 = 2 7. 10j>/ + 2 = -18 
4.-3 = 15 8. 2 £-1 = 9 



154 Fundamentals of High School Mathematics 

9. Throe limes a certain number, increased by 10, 
gives l>. What is the number ? 
10. If twice a certain number is added to 16, the 
result equals the number increased by 6. Find 
the number. 

IX. V+ •>="!->•+ Vr+2. 
o 4 b 

12. 12 -2*= 8. 

13. The sum of two thirds of a certain number and 
three fourths of the same number is — 17. 
Find the number. 

A new kind of equation. 

14. -2*- 12 = 5* -40. 

The equations which you have just solved are of the 
kind in which you can easily see what to do to each side. 
With examples like 14, however, in which both knozvns 
and unknowns occur on each side and which include nega- 
tive numbers on one or both sides, we need special and 
systematic practice. 

Section 67. We need to get "knowns" on one side and 
" unknowns " on the other. Just as clerks in stores always 
place the known weights on one scale pan and the un- 
known weights on the other scale pan, so we, in solving 
equations, always get the known numbers, or terms, on one 
side of the equation, and the unknown terms on the other side. 

Usually we get all the unknown terms on the left side, 
and all the known terms on the right side. Thus, in the 
equation above, 

-2* -12 = 5* -40, 

we do not want — 12 on the left side. Therefore we get 
rid of the known on the left side by adding + 12 to each 






Complete Solution of the Simple Equation 155 

side, giving the equation — 2.;r = 5;r — 28. We also do 
not want the 5x on the right side. Therefore we subtract 
5x from each side, giving the equation — lx = — 28. 
Dividing each side of this equation by — 7, we find that 
* = 4. 

Section 68. Equations should be solved in a systematic 
order. In learning to solve equations which require several 
steps, pupils make many mistakes because their work is not 
arranged in a set order. For some time to come, therefore, 
you will find it very important to use a form like the fol- 
lowing in solving equations : 

Illustrative example. Solve the equation 
- 2 x - 12 = 5 x - 40. 

(1) Adding + 12 to each side, gives 

- 2 x = 5 x - 28. 

(2) Subtracting 5 x from each side, gives 

- 7 x = - 28. 

(3) Dividing each side by — 7, gives 

x = 4. 

(4) Check : Substituting 4 for x to check the result, gives 

- 8 - 12 = 20 - 40. 
_ 20 =-20. 



EXERCISE 72 

Solve each of the following examples, writing out each 
step exactly as in the solution of the illustrative example : 

1. -3*- 8 = 8*- 30 8. 14 = 2j + 20 

2. 5j/-6 = 9j/ + -i2 9. _7tf + 4=+8tf-41 

3. -6£ + H = 2# + 43 10. _2.r-T=-8.r-19 

4. jr- 20 = 50 -ft* 11. +5>-3 = 8^-16 

5. _2r + 10 = 4 12. 5+2jk=0 

6. 10 -8x = -20 13. 10.r + 22 = 12 

7. 6-4j/=2 14. = 4.r + 20 



156 Fundamentals of High School Mathematics 



II. HOW TO SOLVE KQUATIONS WHICH CONTAIN 
PARENTHESES 

Section 69. We saw in the last chapter that parentheses 
were used to indicate multiplication. Thus, to show that 
— 4 is to be multiplied by — 6, we use the parentheses, as 
follows: ( — 4)( — (J). Multiplication is usually indicated 
in this way. Take this example to illustrate the way in 
which parentheses will be used in a practical way : 

A. Illustrative example. Oranges cost 10 cents more per 
dozen than lemons ; the cost of four dozen lemons and two 
dozen oranges is $ 2. What is the price per dozen of each? 

Z?. Solution : 

Let x = no. cents per dozen paid for lemons ; 
then x + 10 = no. cents per dozen paid for oranges; 
then 4x = no. cents paid for 4 doz. lemons; 
and 2(x + 10) = no. cents paid for 2 doz. lemons. 
Therefore 4 x + 2(x + 10) = no. cents paid for both; 
or 4jt + 2(x+ 10)= 200. 

Note the use that is made of parentheses; that is, to show 
that the expression x + 10 must be multiplied by + 2. Per- 
forming this multiplication, 

(1) or removing parentheses, gives 

4x + 2x + 20 = 200. 

(2) Combining terms gives 

6at+ 20 = 200. 

(3) Subtracting 20 from each side gives 

6x = 180. 

(4) Dividing each side by 6 gives 

x = 30 cents per dozen for lemons, 

and therefore 

x + 10 = 40 cents per dozen for oranges. 

(5) Substituting in the original word-statement 30 cents 
and 40 cents respectively for the cost of lemons and oranges 
enables us to check the result : 

4.*. 30+ 2>5.40 = >52 



Complete Solution of the Simple Equation 157 

EXERCISE 73 
PRACTICE IN SOLVING EQUATIONS WHICH CONTAIN PARENTHESES 

Solve and check each of the following equations : 

1. 2(.r + 10)=42 5. 2(*-3) + 3(;r-2) = 8 

2. 5(j - 2)= 15 6. 5 b -f 2(4 - 6)= 32 

3. 3(2 £- 4) = 18 7. -3.r + 6(.r-4)=9 

4. 4^ + 5(.r+2) = 46 a -7£ + 4(2£-3) = 16 

Note that in all the foregoing examples the number be- 
fore the parenthesis has been positive. If negative num- 
bers occur, however, we proceed just the same, remember- 
ing how to multiply a negative number. 

Illustrative example. Solution of an equation involving RE- 
MOVAL OF PARENTHESES. 

9. Sx— 2(2x- 7) = x + 8. 
The expression 2 x — 7 is to be multiplied by — 2. 

(1) Performing this multiplication, or removing parentheses, 
gives 

8jc-4jc + i4 = x + 8. (Why is it +14?) 

(2) Combining terms gives 

4x + 14 = x + 8. 

(3) Subtracting -f x from each side gives 

3x4- 14 = 8. 

(4) Subtracting + 14 from each side gives 

3x=-6. 

(5) Dividing each side by 3 gives 

x=-2. 

(6) Check : Substituting — 2 for x throughout the equation 
gives 

-16-2(-4-7)=-2 + 8. 
- 16 + 8 + 14 = - 2 + 8. 
6 = 6. 



1 58 Fundamentals of High School Mathematics 
EXERCISE 73 (continued) 

10. 5£-3(4-2*)=r2£ + 42 

11. (i(.r-:i)-4(x + 2)=4-.r 

12. 7(6 - 2)- 2(3 + d)= 

13. 4(2^ -5)+ 15 = 3^ + 10) 

14. <)j-8(2j/-4)=6 

Section 70. A difficult form of multiplication. A form 

of multiplication (as shown by parentheses) that gives 
pupils difficulty is the kind represented by — (4 — 5;r) in 
the equation : 

15. 5-2(*-6)=-(4-5;r) 

When no multiplier appears immediately before the 
parentheses, the multiplier 1 is understood. Therefore in 
this case the multiplier is — 1. It is just as though the 
right side of the equation read 

-1(4-5^). 

Illustrative example. 

Therefore the complete set of steps required to solve 
this equation includes: 

(1) Removing parentheses gives 

5 -2x + 12=-4 + 5x. (Whyisit + 5x?) 

(2) Combining terms gives 

-2x + 17=-4 + 5x. 

(3) Subtracting + 17 from each side gives 

-2x=-21+5x. 
(4; Subtracting — 5 x from each side gives 
-7jc = - 21. 

(5) Dividing each side by — 7 gives 

x = 3. 

(6) Substituting 3 for x, throughout the original equation to 
check the result, gives 

5-2(3 - 8) = -(4- 53). 
or, 5 -6 + 12 =-4 + 15. 

11 = 11. 



Complete Solution of the Simple Equation 159 

There are two important and difficult points in this last 
example. First, you should note that in the expression 
5 — 2(x — 6) the — 2 is not to be subtracted from the 5. 
The expression in parentheses must be multiplied 'by — 2. 
Second, if no multiplier is written before the parentheses, 
as in the expression — {\ — bx\ it is understood that the 
multiplier is 1. In this case it is — 1. If there had been 
no sign before the parentheses, as (1 — 5x), the multiplier 
would be understood to be + 1. 

EXERCISE 73 {continued) 

16. lx-{x — 4) =,25 21. 2 £-7(3 -£)=£+ 8 

17. _5 J/ _(2-j)=18 22. 1(2* + 3)= -17 

18. 6x-(x+l)= -2,r + 35 23. —1(6 — 2*)= 43 

19. 5 -2(^-4)= 23 24. -(6-2jt)=24 
20.' 7 - 12(3 - 0)= 31 25. 16 = (2 x + 1) 

A. SUMMARY OF THE STEPS REQUIRED TO SOLVE 
EQUATIONS WHICH CONTAIN PARENTHESES 

Section 71. Look back to the illustrative examples, 9 and 
15, and compare the steps in the solution that is worked out 
for each one with the steps in the solution in eacl) of those 
which you have just worked. You will note that to solve 
such an equation the following steps are always included : 
I. Removing the parentheses {i.e. multiplying). 
II. Combining like terms on each side. 
III. Getting rid of all known terms on one side 

and all unknown terms on the other side. 
IV. Dividing each side by the coefficient of the 
unknown, to give the numerical value of the 
unknown. 
V. Substituting the obtained value in the original 
equation to check the result. 



160 Fundamentals of High School Mathematics 

III. HOW TO SOLVE EQUATIONS WHICH CONTAIN 
FRACTIONS 

Section 72. If, however, we should be given the equa- 
tion 

2(x - 3) _ x - (> = x _ 16 
5 4 2 5' 

we need to add another step ; namely, 

Getting rid of fractions by multiplying each term by the 

most convenient multiplier. 

Illustrative example. 1. Solution of an equation which 
involves getting rid of fractions. 

2(s-3) x-6 = x 16 
5 4 2 5* 

(1) Removing parentheses gives 

2x - 6 s - 6 _s 16 
5 4 2 5 * 

(2) Multiplying each term by the most convenient multiplier, 
20, gives 

20— — -20— =20 5 -20 r 

(3) Reducing fractions gives 

8x-24-5jt + 30 = 10 x- 64. (Why +30?) 

(4) Collecting terms gives 

3x -j-6 = 10jc-64. 

(5) Subtracting 6 from each side gives 

3x = 10 jc — 70. 

(6) Subtracting 10 x from each side gives 

-7x = -70. 

(7) Dividing each side by — 7 gives 

x = 10. 

(8) Substituting the obtained value of the unknown to check the 
result. 



Complete Solution of the Simple Equation 161 

B. SUMMARY OF THE STEPS REQUIRED TO SOLVE 
EQUATIONS WHICH CONTAIN BOTH PARENTHESES 
AND FRACTIONS 

The solution of this example illustrates all the steps that 
are ever included in solving simple equations. The com- 
plete list now includes : 

I. Removing parentheses. 
II. Getting rid of fractions by multiplying "each* 
term by the most convenient multiplier. 
III. Combining like terms. 
IV. Getting rid of all the known terms on one side 

and all the unknown terms on the other side. 
V. Dividing each side by the coefficient of the un- 
known. 
VI. Substituting the obtained value in the original 
equation to check the result. All these steps 
are not required unless the equation includes 
parentheses and fractions. 

Now that we have learned all of the steps that are neces- 
sary in solving simple equations we need to practice so as 
to be very proficient in this work. Nothing is more im- 
portant in high school mathematics. The next exercise 
is included to provide that practice. 

EXERCISE 74 
PRACTICE IN THE COMPLETE SOLUTION OF EQUATIONS 

3(.r-4) = .r + 8 x ,r-4 _ll 

7 3 '25 10 

What step in the above list is omitted m working 
this example ? 



io 2 Fundamentals of High School Mathematics 



3. 



4. 



f>(.r-2)_2(.r + 1) = 7 
3 •") 

2(3.r + 1) (.r+2) = 13 
7 14 14 

3^ 2(6-10) 5J-19 

4 : 5 3 



6. 


.](4.t- + 8)-§(6,r-9)=0 


7. 


1(6* -9)- f(8;r+ 4)= -13 


8. 


5(,-2) 7(2, - 3) _ 12 
3 4 ^ T 4 


g 


2(* + 8) 3(.r-8) -3(2*+ 8) 




7 2 1 


10. 


3(4-,)= 5(6 -3,) 


11. 


x x x — 2 .r 
2 3 6 36 


12. 


2£ 3£ b-b £-37 
5 4 10 20 



TTTi: ALGEBRAIC SOLUTION OF WORD PROBLEMS 

Section 73. Review of important steps in translating 
word problems into algebraic statements. We have taken 
a great deal of time to learn how to solve any kind of 
simple equation because we need to be able to use equa- 
tions skillfully in solving actual problems later. The 
problems as a rule will not be stated for us, in algebraic or 
equational form, all ready for solution. They will be 
stated merely in words. First, therefore, we shall always 
have to translate the word problem into an equation. Be- 
yond this first step the work is the mere solution of the 
equation. 



Complete Solution of the Simple Equation 163 

Our second principal task in this chapter, therefore, 
is to become skillful in translating word problems into 
algebraic form. We learned in our work with Chapter III 
the important steps in translating word problems. Since 
we are to learn in the next few lessons how to translate a 
great many different kinds of word statements, let us re- 
view these steps here : 

First step:. See clearly which things in the prob- 
lem are known and which are un- 
known. 

Second step : Represent one of the unknowns, most 
conveniently the smallest one, by some 
letter. 

Third step : Represent all of the others by using 
the same letter. 

Fourth step : By careful study of the relations 
between the parts of the problem, ex- 
press the word statement in algebraic 
form. 

Sometimes this will mean an equation and sometimes not. 

For the next few lessons, therefore, you will work many 
word problems. The exercises are included to give prac- 
tice in translating many different kinds, so that you will be 
able to use the method in solving any kind that you may 
happen to meet later. For convenience they will be 
arranged by types, examples of the same type being 
studied together. 

I. PROBLEMS IN WHICH A NUMBER IS DIVIDED INTO 
TWO OR MORE PARTS 

Section 74. The solution of a great many problems 
depends upon our being able to separate a number into 



1 04 Fundamentals of High School Mathematics 

two or more parts. For example, if a man has a certain 
sum of money to invest, he may invest part of it in one 
thing, and part in another. The solution of such an ex- 
ample requires that we be able to divide a number into 
two or more parts algebraically. 



EXERCISE 75 
PRACTICE IN DIVIDING A NUMBER INTO TWO OR MORE PARTS 

1. The sum of two numbers is 20. 

(a) Express in algebraic form the second one if 
the first one is 12. 

(b) Express in algebraic form the second one if 
the first one is n. 

(c) Express in algebraic form the fact that the 
second one exceeds the first one by 4. 

2. There are 36 pupils in a mathematics class. 

(a) Express algebraically the number of boys if 
there are 19 girls. 

(b) Express algebraically the number of girls if 
there are n boys. 

(c) State algebraically that there were 6 more 
girls than boys. 

3. A farmer has two kinds of seed, clover seed and 
blue grass seed. If he has 100 lb. of both, 
express : 

(a) the number of pounds of clover seed if there 

were 24 lb. of blue grass seed ; 
{&) the number of pounds of clover seed if there 

were n lb. of blue grass seed ; 



Complete Solution of the Simple Equation 165 

(c) the value of the clover seed {n lb.) at 20 cents 
per pound ; and the value of the blue grass 
seed at 15 cents per pound. 

(d) State by an equation that the value of both 
kinds together was $19. 

4. Divide 20 into two parts such that the larger 
part exceeds the smaller part by 4. 

5. A boy paid 18 cents for 20 stamps ; some cost 
two cents each and the remainder cost three 
cents. How many of each kind did he buy ? 

6. During one afternoon a clerk at a soda fountain 
sold 200 drinks, for which he received $16. 
Some were 5 cents each; the others were 10 
cents each. Find the number of each kind. 

7. A grocer has two kinds of coffee, some selling 
at 30 cents per pound and some selling at 
50 cents per pound. How many pounds of 
each kind must he use in a mixture of 100 
pounds which he can sell for 31 cents per 
pound ? 

Section 75. Need for tabulating the data of word prob- 
lems. Many problems involve so many different state- 
ments that it is practically necessary to arrange the steps in 
the translation in very systematic tabular form. Take an 
example like this : 

John's age exceeds James* s by 20 years. In 15 years he will be 
twice as old as James. Find the age of each now. 

Before we can write this statement in the form of an equa- 
tion we must express in algebraic f oxm four different things : 
(1) John's age now ; (2) James's age now ; (3) John's age 



1 66 Fundamentals of High School Mathematics 

in 15 years; and (4) James's age in 15 years. These 
four facts can best be stated in a table like this : 

(First step) Let it represent James's age now. 

(Second step) Tabulate the data : 

Table 9 





Age now 


Age in 15 years 


Johns age 


n + 20 


H + 20+15 


James's age 


n 


n +i5 



With all the facts expressed in letters we can now state 
the equation which tells the same thing as the original word 
statement ; namely : 

(Third step) n + 20 + 15 = \n + 15). 

We are now ready for the 

(Fourth step) the solution of the equation ; the steps are 
as follows : 

(1) n + 35 = 2^ + 30. 

(2) - n = -5. 

(3) ' » = 5. 

Therefore James's age now is 5, and John's age now 
is n + 20, or 25. 

(4) Check the accuracy of this result thus : 

In 15 years John will be 40 and James will be 
20 ; or John will be twice as old as James, as 
the problem states. 
To be proficient in solving such problems, therefore, we 
first need practice in tabulating such facts as " age now" 
"age some other time," as in this example. Other types 
which involve the same need for tabulation will be taken 
up later. 



Complete Solution of the Simple Equation 167 

PRACTICE IN REPRESENTING RELATIONS BETWEEN 

NUMBERS 

II. PROBLEMS RELATING TO AGE 
EXERCISE 76 

1. A man is now 25 years of age. What expression 
will represent his age: 

{a) 10 years ago ? (c) x years ago ? 

(b) 8 years from now? (d) m years from now? 

2. C is now n years of age. What expression will 
represent his age: 

{a) 12 years from now ? (c) y years ago? 

(b) 7 years ago? (d) m years from now? 

3. A is now x years old. B's present age exceeds 
A's age by 8 years. What expression will repre- 
sent : 

(a) B's present age ? 

(b) the sum of their ages ? 

(c) the age of each 10 years ago? 

(d) the age of each 5 years from now ? 
{e) the sum of their ages in 5 years ? 

4. A is now n years of age ; B is three times as old. 
Express algebraically : 

(a) B's present age ; 

(b) the age of each 4 years ago ; 

(c) the age of each 9 years from now. 

(d) State algebraically that B's age 4 years ago 
was 5 times A's age then. 



i6S Fundamentals of High School Mathematics 

5. A's present age exceeds B's present age by 25 
( years. In 15 years he will be twice as old as B. 

Find their present ages. 

6. C is six times as old as D. In 20 years C's age 
will be only twice D's age 20 years from now. 
What are their present ages ? 

7. A man is now 45 years old and his son is 15. In 
how many years will he be twice as old as his 

son ? 

8. A father is 9 times as old as his son. In 9 years 
he will be only 3 times as old. What is the age 
of each now ? 

9. A's present age is twice B's present age ; 10 
years ago A's age was three times B's age then. 
Find the age of each now. 

III. PROBLEMS BASED ON COINS 

Section 76. Another illustrative type of word problem 
which gives practice in tabulating data and thus in solving 
difficult word problems is the " coin problem." Take this 
example : 

Illustrative example. A man has 3 times as many 

dimes as quarters. 

How many of each has he if the value of both together is 

Sll? 

Here there are four distinct numbers to be expressed, as in 

the case of the age problem: (1) the number of quarters; 

(2) the number of dimes; (3) the value of the quarters in 

terms of a common base (for example, cents) ; (4) the value 

of the dimes in the same base (cents). The steps in the 

solution are clear, therefore, from the following illustrative 

solution : 



Complete Solution of the Simple Equation 169 



(1) Let 

(2) Then 



n = the number of quarters. 

Table 10 





Number 


Value 


quarters 


n 


25 n 


dimes 


3n 


30 n 



(3) 25 n -f 30 n = 1100 cents. 

(4) .-. n = 20, number of quarters. 

(5) 3n = 60, number of dimes. 
(a) Value of the quarters = 85. 

(6) Fa/i/e of the dimes = 9 6. 

Total value = $ 11, as stated in the example. 



EXERCISE 77 



PRACTICE IN EXPRESSING THE VALUE OF VARIOUS NUMBERS OF COINS 

1. Express the value in cents of : 

(a) d dimes ; (d) 4 d half dollars ; 

(b) 3 d quarters ; (e) d dollars ; 

(c) 2 d nickels ; (/) of all the coins. 

2. Express the value in cents of: 

(a) n nickels ; (c) (u + 5) quarters; 

(b) (3 - it) dimes ; (d) (12 - n) half dollars; 

{e) (30 — ri) nickels. 

3. A purse was found which contained nickels and 
dimes, 20 in all. Find the number of each if the 
value of both was 11.60. 

4. I received at a candy counter twice as many dimes 
as quarters, and 6 more nickels than dimes and 
quarters together. How many of each coin did 
I receive if the value of all was $7.50 ? 



170 Fundamentals of High School Mathematics 

5. A debt of $ 72 was paid with 5-dollar bills and 
--dollar bills, there being twice as many of the 
latter as of the former. Find the number of 
each kind of bill. 

6. IS coins, dimes and quarters, amount to $2.25. 
Find the number of each kind of coin. 

7. A cab driver received twice as many quarters as 
half dollars, and three times as many dimes as 
half dollars; in all he had $13. How many of 
each coin did he receive? 

IV. PROBLEMS BASED ON TIME, RATE, AND DISTANCE 

Section 77. In Chapter IV we saw that the motion of a 
train could be represented graphically. Now we shall 
learn how to solve this kind of problem by means of the 
equation. 

EXERCISE 78 

PRACTICE IN SOLVING PROBLEMS BASED ON RELATIONS BETWEEN TIME, 
RATE, AND DISTANCE 

1. Express the distance covered by an automobile 
in 10 hours if its rate is : 

(a) 18 miles per hour; 

(b) 5 miles per hour; 

(c) (r+ 3) miles per hour; 

(d) (2 x — 5) miles per hour. 

2. A train runs for / hours. Express the distance 
it will cover at the rate of : 

{a) 85 miles per hour; 

(b) m miles per hour; 

(c) (r+ 6) miles per hour; 

(d) t miles per hour. 



Complete Solution of the Simple Equation 171 

3. An automobile tourist sets out on a 400-mile 
trip. Express the time required if he goes at the 
rate of : 

(a) 40 miles per hour ; 

(b) 5 miles per hour ; 

(c) (r+ 10) miles per day; 

(d) (2 r — 3) miles per day. 

4. How long will it require to make a trip of D 
miles at the rate of 15 miles per hour ? 5 miles 
per hour ? 

5. At what rate must one travel to go D miles in 
10 hours ? In t hours ? In / + 3 hours ? 

6. A slow train travels at the rate of 5 miles per 
hour; a fast train travels 15 miles more per 
hour. Express : 

(a) the rate of the fast train ; 

(b) the distance passed over by each in 5 hours. 

(c) State algebraically that the two trains to- 
gether traveled 275 miles in 5 hours. 

7. Two trains leave Chicago at the same time, one 
eastbound, the other westbound. The east- 
bound train travels 10 miles less per hour than 
the westbound train. Express : 

(a) the rate of each ; 

(b) the distance traveled by each in 4 hours. 

(c) Form an equation stating that they were 
440 miles apart at the end of 4 hours. 

8. Two trains, 350 miles apart, travel toward each 
other at the rate of 40 and 35 miles per hour, 
respectively. 



[72 Fundamentals of High School Mathematics 

(a) Express the distance traveled by each in / 

hours. 
(6) Form an equation stating the fact that the 

trains met in / hours. 
9. Make formulas for d, for t, and for r, that can 
be used in any problem based upon uniform 
motion. 

10. Illustrative example. Two bicyclists, 200 miles apart, 
travel toward each other at rates of 12 and 8 miles per hour 
respectively. In how many hours will they meet ? 
(1) Let t represent the number of hours until they meet. 



(2) 



Table 11 





Time in 
hours 


Rate per hr. 
in miles 


Distance in 
miles 


For slow one 
For fast one 


t 
t 


8 
12 


8t 

12 1 



(3) Then 

(4) 



8 t + 12 t = 200. 
.-. * = 10. 



11. Two men start from the same place, one going 
south and the other going north. One goes 
twice as fast as the other. In 5 hours they are 
1^0 miles apart. Find the rate of each. 

12. An eastbound train going 30 miles per hour left 
Chicago 3 hours before a westbound train going 
36 miles per hour. In how many hours, after 
the westbound train left, will they be 519 miles 
apart ? 



Complete Solution of the Simple Equation 173 

13. A bicyclist traveling 15 miles per hour was over- 
taken 8 hours after he started by an automobile 
which left the same starting point 4 J hours later. 
Find the rate of the automobile. 

14. A starts from a certain place, traveling at the 
rate of 4 miles per hour. Five hours later B 
starts from the same place and travels in the 
same direction at the rate of 6 miles per hour. 
In how many hours will B overtake A ? 

V. PROBLEMS INVOLVING PER CENTS 

Section 78. Many problems involving per cents may be 
solved by algebraic methods. 

EXERCISE 79 
PRACTICE IN SOLVING PERCENTAGE PROBLEMS 

1. What does 10% mean ? 5% ? r%J 

2. Indicate 4 % of $600 ; 5 % of 1275. 

3. Express decimally 5 % of / ; 8 % of c ; 6^ % of b. 

4. A man paid c dollars for an article. He sold it 
at a gain of 25 %. Express : 

(a) the gain in dollars ; 

(b) the selling price. 

(c) State algebraically that he sold the article 
for $2.50. 

5. A merchant sold a suit for $25, thereby gaining 
25%. If the cost is represented by c dollars, 
what will represent : 

(a) the gain in dollars ? 

(b) the selling price in terms of c ? 

(c) State algebraically that the selling price was 
$25. 



174 Fundamentals of High School Mathematics 

6. Solve each of the following equations: 

(*) .20 x= 180 

(b) x + Mx = 3.18 

(c) c + .10c = VX r > 

(d) vi - Xhm = 21.25 

(e) p + .04/ = 520 
if) x- .50 x = 18.75 
\g) 2 - Sx- ,5x= 7 
(h) 1.15 x- \x= 1000 

7. Find the cost of an article sold for $ 156 if the 
gain was 10 %. (Use c for the cost.) 

8. What number increased by 66| % of itself 
equals 150 ? 

9. After deducting 15 % from the marked price of 
a table, a dealer sold it for % 21.25. What was 
the marked price ? 

10. A dealer made a profit of $ 3690 this year. 
This is 18 % less than his profit last year. Find 
his profit last year. 

11. A number increased by 12.5 % of itself equals 
213. What is the number ? 

12. A shoe dealer wishes to make 25 % on shoes. 
At what price must he buy them in order to 
sell them at $4.50 per pair? 

13. A furniture dealer was forced to sell some dam- 
aged goods at 14 % less than cost, and sold 
them for $ 129. How much did they cost ? 

14. A man sold a suit of clothes for % 30.25. What 
per cent did he gain if the clothes cost him 
$ 25 ? 



Complete Solution of the Simple Equation 175 

VI. INTEREST PROBLEMS 

Section 79. Many interest problems can be more easily 
solved by algebraic equations than by the methods of 
arithmetic. 

EXERCISE 80 

1. Express the interest on $150 at 5 % for 1 year; 
for 3 years ; for / years. 

2. Express the interest on P dollars at 6 % for 1 
year ; for 3 years ; for / years. 

3. Express the simple interest on #500 for 1 year 
at r °/o ; for 4 years. 

4. A man borrowed a certain sum of money at 6 %. 
Express : 

(a) the interest for 2 years. 

(b) State algebraically that the interest for three 
years was $48. 

5. What principal must be invested at 6 % to yield 
an annual income of $ 57 ? 

6. For how many years must $ 2800 be invested 
at 7 % simple interest to yield $ 833 interest ? 

7. What is the interest on P dollars at r% for / 
years ? 

8. A man invests part of 1 1000 at 4 %, and the 
remainder at 6 %. If x represents the number 
of dollars invested at 4 %, express : 

(a) the annual income on the 4 °j investment; 

(b) the amount of the 6 % investment ; 

(c) the annual income on the 6 % investment. 

(d) State algebraically that the annual income 
on the 4 % investment exceeds the annual 
income on the 6 °/ investment by $ 20. 



176 Fundamentals of High School Mathematics 

9. Part of $1200 is invested at 5% and the re- 
mainder at 7 %. The total annual income from 
the two investments is $67. What was the 
amount of each investment ? 

10. Ten thousand dollars' worth of Liberty Bonds 
yield an annual interest of $ 370. Some pay 
o\ %, and the remainder pay 4 %. Find the 
amount of each kind of bond. 

11. A 5 % investment yields annually $5 less than 
a 4 % investment. Find the amount of each 
investment if the sum of both is $ 800. 

VII. PROBLEMS CONCERNING PERIMETERS AND AREAS 

Section 80. The following examples are based on 
squares and rectangles. 

EXERCISE 81 

1. The length of a rectangle exceeds twice its width 
by 12 in. Represent its width by w. 

(a) Make a drawing to represent it. 

(b) Express its length. 

(c) Express its area. 

(d) Express its perimeter. 

(e) State that its perimeter is 84 in. 

2. The length of a rectangle is 9 in. more, and the 
width is 6 in. less, than the side of a square. 

(a) Make a drawing for each. 

(b) Express the dimensions of the square. 

(c) Express the dimensions of the rectangle. 

(d) Express the perimeter of the rectangle. 
(c) Express the area of the rectangle. 

(/) State algebraically that the sum of the perim- 
eters is 168 in. 



Complete Solution of the Simple Equation 177 

3. The base of a triangle exceeds its height by 10 
inches. 

{a) Make a drawing for the figure. 

(b) Express its base and height. 

(e) Express its area. 

(d) State that its area is equal to the area of a 

rectangle whose dimensions are 8 in. and 

5 in. 

4. The length of a rectangle is 4 feet more, and 
its width is 2 feet less, than a square whose perim- 
eter is P inches. Express : 

(a) the side of the square ; 

(b) the dimensions of the rectangle; 

(c) the perimeter of the rectangle. 

(d) Find the value of P if the perimeter of the 
rectangle is 44 inches. 

VIII. PROBLEMS BASED ON LEVERS 

Section 81. A teeter board is one form of lever. The 
point on which the board rests or turns is the fulcrum; 
the parts of the board to the right of and to the left of the 
fulcrum are the lever arms. 




If a boy at A, who just balances a boy at B, moves to 
the left while B remains stationary, it is clear that the left 
side goes down. But if the boy at B moves closer to the 



178 Fundamentals of High School Mathematics 

fulcrum while A remains stationary, then A goes down. 
It is also clear that boys of unequal weight cannot teeter 
unless the heavier boy sits closer to the fulcrum. There 
is a mathematical relation between the weight on the lever 
arm and its distance from the fulcrum. Two boys will 
balance each other when the weight of one times his dis- 
tance from the fulcrum is equal to the weight of the other 
times his distance, or in general, when 

weight times distance on one side equals weight times 
distance on the other side. 

This law or relation may be tested by placing equal coins 
at different positions on a stiff ruler balanced on the edge 
of a desk. Try this experiment. See whether 2 pennies 
placed 4 inches from the fulcrum (at the center of the 
lever) will balance 1 penny placed 8 inches from the ful- 
crum. See whether 6 pennies placed 2 inches from the 
fulcrum will balance 3 pennies placed 4 inches from the 
fulcrum. 

Thus, to make a lever balance, the product of weight 
and distance from the fulcrum on one side must equal the 
product of weight and distance from the fulcrum on the 
other side. 



EXERCISE 82 
Problems based on levers. Make a drawing for each. 

1. John weighs 80 lb. and sits 4 ft. from the ful- 
crum. Where must Robert sit if he weighs 90 lb. ? 

2. A, weighing 120 lb., sits 4* ft. from the fulcrum, 
and balances B, who sits 5 ft. from the fulcrum. 
What is B's weight ? 



Complete Solution of the Simple Equation 179 

3. A hunter wishes to carry home two pieces of 
meat, one weighing 40 lb. and the other 60 lb. 
He puts them on the ends of a stick 4 ft. long 
and places the stick across his shoulder. Where 
must the fulcrum (his shoulder) be placed to 
make the weights balance ? 

4. Two children play teeter, one on each end of a 
board 9 ft. long. Where must the fulcrum be if 
the children weigh 60 and 80 lb. respectively ? 

5. Could three children teeter on the same board ? 
How ? 

6. A and B sit on the side of the fulcrum. A 
weighs 100 lb. and sits 5 ft. from the fulcrum ; B 
weighs 80 lb. and sits 3 ft. from the fulcrum. 
Where must C sit to balance the other two, if he 
weighs 150 lb. ? 

7. Show that — \ = _2 is the formula or law for a 

balance on a teeter board. Are the weight and 
distance directly proportional ? 

SUMMARY 

This chapter has taught all the steps involved in solv- 
ing a simple equation : 

1. Removal of parentheses. 

2. Getting rid of fractions. 

3. Collecting terms on each side. 

4. Getting rid of known terms on one side and un- 
known terms on the other side. 

5. Dividing each side by the coefficient of the un- 
known. 



1S0 Fundamentals of High School Mathematics 

6. Checking by substituting the obtained value of 
the unknown in the original equation. 

Many kinds of word problems have been solved. Tabu- 
lating the information or data of such problems is a great 
help in solving them. A systematic method always pays big 
dividends in any kind of work. 

EXERCISE 83 
MISCELLANEOUS PROBLEMS 

1. A grocer has two kinds of tea, — some worth 
60 f! per pound and some worth 75 per pound. 
He has 20 lb. more of the former than of the 
latter kind. How many pounds of each kind 
has he, if the value of both kinds is $45.75 ? 

2. I bought 45 stamps for $1.05. If part of them 
were 2-cent stamps and part 3-cent stamps, 
how many of each did I buy ? 

3. The sum of the third, the fourth, and the eighth 
parts of a number is 17. What is the number? 

4. John has I as many marbles as Harry. If John 
buys 120 and Harry loses 23, John will then 
have 7 more than Harry. How many has each 
boy ? 

5. A clerk spends \ of his yearly salary for board 
and room, \ for clothes, £ for other expenses, 
and saves $880. What are his annual expenses? 

6. A father left one third of his property to his 
wife, one fifth to each of his three children, and 
the remainder, which was $1200, to other rela- 
tives. Find the value of his estate. 



Complete Solution of the Simple Equation 181 

7. Ten years ago A was one third as old as he is 
at present. Find his age now. 

8. Evaluate the formula C='-± — • ' if 7^=20. 

9. A merchant bought goods for §500, and sold 
them at a gain of 5%. What was the selling 
price ? 

10. If in problem 9 the merchant had sold the 
goods at a gain of x per cent, what would have 
been the selling price ? 

11. A 6-foot pole casts a shadow 4 J ft. in length. 
At the same time how long is the shadow of an 
8-foot pole ? 

12. The ratio of two numbers is f. Find each 
number if their sum is 56. 

13. Two numbers differ by 70 ; the ratio of the 
larger to the smaller is -|-. Find each number. 

Cj 

14. In Fig. 97, ZC= 90°, 
A A = 37°, and 
AC=2±. Find AB, 
BC, and Z B. 

Fig. 97 

15. In general, which is the larger, the sine or the 
tangent of an angle ? Show by a drawing. 

16. The highest office building in the world (the 
Woolworth Building, New York City) casts a 
shadow 1240 ft. long at the same time that a 
boy 5 ft. tall casts a shadow 8 ft. long. What 
is the height of the building ? 




[82 Fundamentals of High School Mathematics 

17. The table below gives the annual cost of pre- 
mium per $1000 life insurance, at various ages. 



Age 


21 


25 


30 


35 


40 


45 


50 


Premium 


18.25 


20.04 


22.60 


26.40 


30.50 


3610 


45.20 



Show this graphically. Measure age along the 
horizontal axis. 



CHAPTER XI 

HOW TO SOLVE EQUATIONS WHICH CONTAIN TWO 
UNKNOWNS 

I. GRAPHICAL SOLUTION 

Section 82. Importance of skill in drawing the picture 
or graph of an equation. In Chapter IV we learned how 
to represent and to determine the relationship between 
quantities that change together. Three methods of doing 
this were studied : (1) the tabular method ; (2) the graphic 
method; (3) the equational or formula method. One of 
the most important facts for us to recall is that the graph 
and the equation tell exactly the same tiling. For exam- 
ple, on 'page 51, the line BC and the equation C= .12n 
tell exactly the same thing. Any information that you 
get from the equation you can also get from the graph. 
Furthermore, relationships can be seen more easily from 
graplis than from tables or equations. For these reasons, 
and since much of our later work in mathematics makes 
use of graphic methods, we need to be skillful in drawing 
the line which stands for an equation. 

Section 83. We need to know how to locate or to "plot " 
points.. But a line may be regarded as a series of points. 
Thus, to represent or locate a line we have to locate a 
series of its points. It happens that much of our elemen- 
tary work, furthermore, deals with straight lines. This 
kind of line, clearly, can be fully determined by locating 
any two of its points. 

HOW TO LOCATE OR PLOT POINTS 

Thus, we see that the important thing in "graphing" 
is how to locate, or to represent, points. In every graph 
that you have already constructed you have had to locate 

183 



184 Fundamentals of High School Mathematics 

joints through which to draw the line. For example, in 
constructing a cost graph it is necessary to locate several 
points representing the cost of different numbers of units 
of the article. Let us study more carefully how points 
are located. 

Section 84. How points are located on maps. (1) Points 
are located on maps by means of latitude and longitude. 
Any point on the earth's surface is definitely located by 
stating its distance cast or west of the prime meridian, and 
its distance north or south of the equator. 

Thus, to the nearest degree, the location of New York 
is 74° W. and 41° N. because it is 74° west of the prime 
meridian and 41° north of the equator. Similarly, the 
position of Chicago is 88° W. and 42° N. ; that of Paris, 
2° E. and 49° N. 

(2) This same method is used by many cities in number- 
ing their houses. Two streets, which make right angles 
with each other, are selected as reference streets. Any 
house or building is completely located, then, by stating 
the number of blocks it is east or west, and north or south, 
of these reference streets. 

Section 85. How points are located on drawings. By a 
method similar to that above, we locate points on 'paper. 
Instead of using the equator and the prime meridian as 
our reference lines, we take two lines, — for convenience, 
one horizontal and the other vertical, — which make a right 
angle with each other. Any point may be located, then, 
by stating its distance to the fight of, or to the left of, the 
vertical reference line ; and its distance above or below the 
horizontal reference line. 

Thus, in Fig. 98, point A is 2 units to the right of, 
and 1 unit above, the reference lines ; point B is 2 units 
to the left of, and 3 units below, the reference lines ; point 
C is 8 units to the left of, and 2 units below, the reference 



Solving Equations with Two Unknowns 185 



-x 

































D 












































A 

















































c 






















B 

























































X 



-Y 

Fig. 98 

lines ; and point D is units to the right or left of, and 
3 units above, the reference lines. 

Section 86. The point from which distances are meas- 
ured : The origin. The point in which the two axes meet, 
or their intersection point, is called the origin. It is the 
point from which we measure distances, either way. The 
origin is usually lettered with a capital O, as in Fig. 98. 

Section 87. How distances are distinguished from each 
other. It would be laborious to state that a particular 
point is "to the right of" or "to the left of "some refer- 
ence line, each time we refer to it. To avoid this, it has 
been agreed to call distances to the right of the Y-axis 
positive, and distances to the left of the Y-axis negative. 
Similarly, distances above the X-axis are positive, and dis- 
tances below the X-axis are negative. It is very important 



1 86 Fundamentals of High School Mathematics 



to remember these facts because we use them so frequently 
in graphic work. 

Thus, in Fig. 98, the position of point A is described by 
the numbers + 2 and +1, or by (2, 1). This means that 
point A is 2 units to the right of the F-axis, and 1 unit 
above the .Y-axis. Similarly, the position or location of 
point B is described by the numbers — 2 and — 3, or by 
( — 2, — 3); this means that point B is 2 units to the left of 
the F-axis and 2 units above the ^Y-axis. In the same 
way, the position of point C is described by the numbers 
— 3 and — 2, or (— 3, — 2); this means that point C is 3 
units to the left of the F-axis and 2 units below the JYaxis. 

At this time the student should note that in stating the 
location of a point, its distance to the right of, or to the left 
of, the F-axis is always given before its distance above or 
below the X-axis. This is done to avoid confusion. That 
is, the x-distance is always first, the ^/-distance second. 

Section 88. Plotting a point. By "plotting a point" we 
mean the locating, on cross-section paper, of a point whose 
x -distance and j/-distance are known. 






Thus, to plot A, 
whose ^-distance is 
-f 3 and whose ^-dis- 
tance is + 4, usually 
written (3, 4), means 
to locate on the cross- 
section paper a point jyr 
3 units to the right 
of, and 4 units above, 
the origin, as in Fig. 
99. In the same 
way, the point ( — 2, 1) 
is point B on the 
graph. 



+Y 















































~lr >" 














JJ3_ 








1r 












.4; 




-X\ 




•r. 











































































■X 



Fig. 99 



Solving Equations with Two Unknowns 187 

EXERCISE S4 
PRACTICE IN PLOTTING POINTS 

1. The ^--distance of a point is + 3, i.e. it is 3 units 
to the right of the vertical axis. Is it definitely 
located ? Why ? 

2. The j'-distance of a point is — 4, i.e. it is 4 units 
below the horizontal axis. Is it definitely located? 
Why? 

3. A certain point is on both axes. What are its x- 
and j-distances ? 

4. Plot the points whose position is determined by 
the following : (4, 2), (5, 6), (- 3, 2), ( - 4, - 1), 
and (-6, 2). 

5. Plot the following: (2, 8), (3, 7), (4, 6), (5, 5), 
(6, 4), (8, 2), (10, 0). 

6. Plot the following : (12, - 2), (15, - 5), (18, - 8), 
(10, 0), (5, 5). 

7. Plot the following: (21 3), (If, 5), (-21, 3). 

HOW TO DRAW THE GRAPH OF AN EQUATION WHICH 
CONTAINS TWO UNKNOWNS 

Section 89. The picture of an equation. Now that we 
have learned how to locate, or plot, points, we come to the 
main purpose of the chapter : to shozv Jiozv equations can be 
solved grapliically. 

First illustrative example. Let us take an equation 
which contains two unknowns, such as, 

In this equation the value of y changes as the value of x 
changes. Clearly, the value of y depends upon the value of x. 
For example, if x = 1, then y — 5 ; if x = 2, then y = 7, etc. 
A table will help to show this relation between the unknowns, 
x and y. 



[88 Fundamentals of High School Mathematics 



Table 12 



If x equals 


1 


2 


3 


4 


5 


O 


-1 


-2 


-3 


-4 


-5 


then y equals 


5 


7 


9 


11 


13 


3 


1 


-1 


-3 


-5 


-7 



If we select any particular value of x, and the corresponding 
value of y which accompanies it, such as 1 and 5, or 2 and 7, 
we may think of them as completely describing the position 
of points on a graph. Thus, (1, 5), (2, 7), (3, 9), etc., 
definitely locate the position of the points. Plotting these points 
with respect to an X- and F-axis, we get a series of points, 
such as Fig. 100. By joining these points we obtain a straight 
line, which is the picture or the graphical representation of the 
equation y — 2x + 3. 



+Y 



-x- 









*-r- T 




22 




±tt 




-e07 




Z^ f 




-Z+ 




A v 




7 




T 




I 




t 


/ 




T 




/ 


_4T 


ta 


Ti 


/ J 




t 




/ 




t 




/ 




f 




4 




7 









-Y 

Fig. 100 



Solving Equations with Two Unknowns 189 

Second illustrative example. For a second example, 
let us consider an equation in two unknowns which shows 
that the sum of two numbers is always 10, such as x -f y = 10. 
It is clear that one number, x, might be 2, and if so, that y 
must be 8 ; or that x might be 4, and if so, the other number, y, 
must be 6. Thus the two unknowns, x and y, may have many 
different values. A table helps to show this. 

Table 13 



If x equals 


l 


2 


3 


4- 


O 


-1 


-2 


-3 


then y equals 


9 


8 


7 


6 


10 


11 


12 


13 



Now we may think of any pair of related numbers, such as 1 
and 9, or 2 and 8, as describing the position of points on this 
line. Thus, (1, 9), (2, 8), (3, 7), (4, 6), etc., show the location 
of points on the graph. Plotting these points, we have 
Fig. 101. By joining these points we obtain in this illustra- 
tive example the "graph"' or picture of the equation 
x + y = 10. Expressed in another way, we have represented 
graphically the relation between two numbers whose sum is 
always 10. 

+Y 



-X- 



s__ 




_^,_ 




s 


s 




-V^ 




v jL 




- 5^ 




^^ 




S3I 




sL 




Ia -O-'' 


n 


-\ -fc 


V 


_\ 




<v7 































-Y 



Fig. 101 



joo Fundamentals of High School Mathematics 

EXERCISE 85 

PRACTICE IN REPRESENTING GRAPHICALLY EQUATIONS WHICH CONTAIN 

TWO UNKNOWNS 

1. In the equation y = 2x + 5, find the value of y 
when x == 1 ; when x = 2 ; when x = 4 ; when 
x =s ; when x = — 3. Make a table similar to 
the one above, showing these related pairs of 
numbers. 

2. In the equation x = y — 4, find the value of x 
when y = ; when y — 1 ; when j = 4 ; when 
j = 6 ; when j/ = — 2 ; when y = — 6. Tabu- 
late. 

3. Plot the equation given in Example 2. 

4. Plot the equation x + y = 6. Hint : First tabu- 
late related values of x and j/. Use only four 
points. 

5. Plot, or graph, y = 5 + x. 

6. Plot 2;r+ y = 6, orj = 6 - 2*. 

7. Graph x — 2 j/ = 5, or x = 2y + 5. 

8. Plot 3;r — y = 8, or j = 8.r — 8. 

Section 90. An easier method of plotting a line. A 

straight line is definitely determined or located if any two 
of its points are known. If these points are not too close 
together, they fix the plotted position of the line just as 
accurately as eight or ten points. Therefore, in plotting a 
straight line, it is sufficient to plot only two points, unless 
they are quite close together. 

The easiest points to plot are those on the axes ; that is, 
the points where the line cuts the .r-axis and the jj/-axis. 
By referring to Fig. 101, or to the graph of any line, you 



Solving Equations with Two Unknowns 191 

will see that the x-distance of the point in which the line 
acts the vertical, or y-axis, is alzvays 0, and that the y-dis- 
tance of tJie point in which the line acts the horizontal, or 
x-axis, is alzvays 0. Thus, if we let x be in any equa- 
tion, such as 2x—y = 6, we find the point in which the 
line cuts the j/-axis. If x is in 2x—y = 6, we see that 
y equals — 6, which shows that the line cuts the y-axis at a 
point (0, —6); that is, 6 units below the origin. In the 
same way, if we let y be in any equation, we find the 
point in which the line cuts the ^r-axis. In this particular 
equation, 2x—y = G, if y is 0, then x is 3, which shows 
the point in which the line 2x — y= 6 cuts the ^r-axis. 

This shorter method requires only the following brief 
table : 

Table 14 



X equals 





? 


y equals 


? 






EXERCISE 86 

1. If x = 0, what is y in the equation 2 x +y = 8 ? 
What is x if y = ? From these two sets of 
values for x and y, plot the equation. 

2. Given 4,r — 2j/ = 8. Plot by finding where the 
line cuts the axes. 

3. Where does the graph of 5x -\-2y = 10 cut the 
jr-axis ? thej/-axis? 

4. Where does the graph of 2x — 3y = — 6 cut 
the ;r-axis ? the j/-axis ? Plot. 

5. Gr aph 2\ vX + y = 5. 



ig2 Fundamentals of High School Mathematics 



HOW TO SOLVE GRAPHICALLY EQUATIONS WITH TWO 

UNKNOWNS 

Section 91. When is an equation with two unknowns 
solved? In equations with only one unknown, such as 
3 x + 4 = 19, we found that there was only one value for x 
which would satisfy the equation; namely, x = b. If we 
substitute 5 for x in this equation, giving 15 + 4 = 19, we 
find that 5 satisfies the equation. Any other number sub- 
stituted for x would not " satisfy the equation." 

But now consider an equation which has two unknowns, 
such as 

;r+j=8. 

Here we see that x might be 3 and y would be 5 ; or x 
might be 6 and y would be 2 ; or x might be 10 and y 
would be — 2. Thus, there are a great many sets of 
values of x and y which could satisfy the equation 

x+y = S. This will be made clear as you work the fol- 
lowing examples. 

EXERCISE 87 

1. Give four sets of values of x and y that will 
satisfy the equation x—y= 6. 

2. Will x = 4} 2 and y = 3 satisfy the equation 
\x — y = 15 ? Does x = 5 and j/=4 satisfy it ? 

3. If the equation x+y = 8 is plotted, would the 
points (5, 3) lie on the line representing the 
equation ? (3, 4) ? (1 0, - 2) ? (1, 7) ? 

4. Does the graph of the equation 2x — y= 7 pass 
through the point (5, 3) ? (4, 2) ? 



Solving Equations with Two Unknowns 193 

We have shown that an equation with two unknowns 
is solved when a set of values for the unknowns is found 
which satisfies the equation. 

Section 92. Linear equations. The fact that the graph 
of an equation which contains two unknowns, each of the 
first degree {i.e. no squares or cubes), is always a 
straight line, has led to the lame linear equations. Thus, 
2x +y = 5, x + 5 = 10, etc., are linear equations. 

TWO LINEAR EQUATIONS MAY BE EASILY SOLVED BY 
PLOTTING THEM ON THE SAME AXES 

Section 93. It is a very common problem in mathe- 
matics to have to find one set of values which will satisfy 
each of two equations having two unknowns. For ex- 
ample, what single set of values will satisfy each of these 
equations ? 

\x+y = 

-J' = ~ 

It is clear that x = 6 and y = 2 or (6, 2) will satisfy the 
first equation, but not the second one ; in the same w T ay 
x = i and 7 = 4 or (4, 4) satisfies the first equation, but 
not the second one ; x = 6 and y = 5 satisfies the second 
equation, but not the first one. 

OUR PROBLEM IS TO FIND ONE SET OF VALUES THAT 
WILL SATISFY BOTH EQUATIONS 

This can be done, graphically, by plotting both equations 
on the same axes, because in that way we can find a point 
common to the two lines ; that is, the point in which two 
lines intersect. The coordinates of this point will satisfy 
both equations. Figure 102, on the following page, shows 
both equations plotted on the same axes. Note that the 



lq4 Fundamentals of High School Mathematics 



* - 'V 


-5> ■ JL-L - 


-^ -At 


K - zl 


^r-tS-' 


5?7 


*cr 


/^ 


r s r 


j S _ _. 


-fi- -/ v»fc 


^ 7 Sp* 


r S*^ 


I Kxl- 


-T 5Z 


7 S 


r^ 


/ 


_/ 





Fig. 102 



two lines intersect at the point (5, 3). This point of inter- 
section of the two lines gives a single set of values, x = 5 
and y — 3, which satisfies both equations. (Show that x = 5 
andj= 3 checks for each of the equations.) 



EXERCISE 



Find a set of values that will satisfy each of the follow- 
ing pairs of equations, by finding the intersection point of 
their graphs : 



1. 



2. 



4. 



5. 



x + y = 6 
2x-y = 3 

y=2x+Z 

y = x+l 
\x — y = 5 
\2x+y = 7 

x _ 3 j, = _ 2 

j/ = .r + 6 



6. 



9. 



\ a -2b = -5 

x+y= 10 
3.r + 3j/ = 6 
„v = j/ + 6 

_r + j = 1 
x — y = 5 



Solving Equations with Two Unknowns 195 



+Y 











->r- 




^ 




^,- 




ST 




s± : 




,_ In 




V St 




JS ^ 




5 1 v 




"^ V- 




^ X 




T ■ s 


K 


.v. • ^t Xii- 


--X- . . -4.TT 


A ISji _ 


\ I A 


±t - 


: s££ 


^ 




K ^ 


1 N Sr s 


^ 


fr '» '5? 




s V> ^ 




or. ^ 




"x^ ^ 




- ^V 




1 — 1 — 1— J — 1 1 1 1 1 1 1 



-Y 



Fig. 103 



Section 94. Equations whose graphs are parallel lines, 
i.e. inconsistent equations. Figure 103 shows the graphs of 
the equations represented below. 

(1) \x+y = ± 

(2) \2x + 2y = -6 

Note that the lines do not intersect, but are parallel. What 
single set of values of x and y will satisfy each of these 
equations ? Evidently there is none, for they have no point 
in common. Such equations are generally called incon- 
sistent, to distinguish them from the kind that are satisfied 
by some set of values. The latter kind, those whose 
graphs intersect, are often called simultaneous equations. 



196 Fundamentals of High School Mathematics 

EXERCISE 89 

Graph each of the following pairs of equations to 
determine which pairs are inconsistent and which are 
simultaneous : 

y — *==4 3 jx+y = 4; 



{ x — 6 = y \ x — y = 6 

l2j/ + 10 = 4.r ' \x + y=12 

SUMMARY 

This chapter should make it clear that : 

1. An equation may be plotted or graphed by locat- 
ing a series of points, the x and y values of which 
will satisfy the equation. 

2. Two equations are solved graphically by finding 
the x and y values of the point of intersection of 
their lines. 

REVIEW EXERCISE 90 

1. From the sum of — 6 and + 10 take — 8. 

2. The product of two numbers is — 40 y; one of 
them is + 10. What is the other ? 

3. State the four principles, or axioms, used in solv- 
ing equations. Illustrate in solving the equation 
by -8 =+2y- 50. 

4. If A = 4:X + Sy and B = ix — 3y, what does 
A + B equal ? What does A - B equal ? 

5. Does -- = '-^— ? Does - = — ? Does - = — ? 

7 7-2 b be 2 10 

State the principle involved in these examples. 



Solving Equations with Two Unknowns 197 



Age 


Boys who leave school at the 
age of 14 earn weekly wages 
as indicated 


Boys who leave school at the 
age of 18 earn weekly wages 
as indicated 


14 


$ 4.00 





16 


5.00 





18 


7.00 


f 10.00 


20 


9.50 


15.00 


22 


11.00 


20.00 


24 


12.00 


24.00 


25 


13.00 


30.00 



6. Studies have been made to determine the money 
value of a high school education. The table 
above shows the average weekly earnings for 
boys who leave school at the age of 14, and for 
those who remain in school until they are 18 
years old. 

Graph the earnings for each class of boys on the 

same axes. Measure age along the horizontal 

axis. 

Interpret the graph. If a boy knew that he 

would live to be only 25 years old, would it pay 

him, in dollars, to go to high school ? How 

much ? 

7. The areas of two circles are directly proportional 
to the squares of their radii, Compare their 
areas if the radius of one circle is 3 times the 
radius of the other. 

8. In solving a particular problem, how do you tell 
whether to use the sine, cosine, or tangent? 
Illustrate by specific examples. 



CHAPTER XII 

HOW TO SOLVE EQUATIONS WITH TWO UNKNOWNS 

(Continued) 

II. SOLUTION BY ELIMINATING ONE UNKNOWN 

Section 95. The need for a shorter method of solving 
equations with two unknowns. In the previous chapter we 
saw that equations with two unknowns can be solved by 
graphic methods. The exclusive use of that method, how- 
ever, would require a great deal of time, and would 
necessitate that we have cross-section paper at all times. 
Fortunately, there is a sJwrter method which can be used. 
This is a method by which we eliminate one of the 
unknowns. By eliminating or getting rid of one of the 
unknowns, we obtain an equation with only one unknown. 
The following illustrative examples will explain the differ- 
ent ways by which one of the unknowns is eliminated. 
This chapter will show three methods of elimination. 

I. ELIMINATION BY COMPARISON; THAT IS, BY 
EQUATING VALUES OF ONE OF THE UNKNOWNS 

Section 96. Equating values of one of the unknowns. 

This method is illustrated by the following example: 



First illustrative example. Find the value of 


x and of y 


in the following equations : 

f*+ 2/ = 10, 
jx-3y = -6. 

Solution : 


(1) 
(2) 


Solving for x in equation (1), 

x = 10 - y 


(3) 


and in equation (2), 

x = 3 y - 6. 


(4) 


Comparing or equating the values of x, 

3 y- 6 = 10- y, 


(5) 


which gives 4 y = 16, # 
or y = 4. 


(6) 
(7) 



Solving Equations with Two Unknowns 199 

Substituting 4 for y in (1) or (2) gives 

jc = 6. 

Checking in equations (1) and (2), 

6 + 4 = 10. 
6 -12 ==-6. 



Elimination by comparison is based upon a fact which 
was illustrated in the graphical solution of equations with 
two unknowns ; namely, that at the point of intersection of 
the two lines, the value of x in one equation is the same as 
the value of x in the other equation, and the value of y in 
one equation is the same as the value of y in the other. 
For this reason, we may form an equation by equating, or 
placing equal to each other, the two values of x y which 
were 10— y and 3j — 6. This process gets rid of, or 
eliminates, the unknown x and gives us an equation with 
only one unknown, y. 

Second illustration of the method of eliminating one 
unknown. The same results could have been obtained by 
finding the value of y in each of the two given equations : 

jc+ y = 10. (1) 

x-Zy=-Q. (2) 

From (1), y = l0-x. (3) 

From (2), -3jf=— 6-x, (4) 

Comparing the values of y 10 — x = ^ « (6) 

3 

Multiplying by 3, 30 - 3 x = 6 + x, (7) 

or -4x=-*, (8) 

or x = S. (9) 

Substituting in (1) or in (2), y = 4. (10) 



200 Fundamentals of High School Mathematics 

EXERCISE 91 

PRACTICE IN ELIMINATING ONE OF THE UNKNOWNS BY COMPARING, OR 
EQUATING, ITS VALUES AS OBTAINED FROM THE TWO EQUATIONS 

Solve and check each of the following : 

* = 2^-3 f.r-3/ = 



\x= by -21 4 * \±s+t=l± 

fr-j/=10 5 <2b + 3c = 6 



j,r = 16-2j/ ' \b = 5c + l6 

[y + 2x=12 ( X + E>j/ = 1 

3 * [5^ + ^ = 42 6 * |2.r+6j/ = -2. 

7. The sum of two numbers is 14; the larger ex- 
ceeds the smaller by 2. Find each number. 

8. Twenty coins, dimes and nickels, have a value 
of $1.70. Find the number of each. 

9. A boy earns $ 2 per day more than his sister; 
the boy worked 8 days and the girl worked 6 
days. Both together earned $44. What did 
each earn per day ? 



10. 



11. 



1^ + ^=10 12 * \§ x = u-y 

2;r + 3j = 5 [j/ = 2.r-10 

Zx-y=2 13 * [x=2y-U 



II. ELIMINATION BY SUBSTITUTION; THAT IS, BY SUB- 
STITUTING THE VALUE OF X FROM ONE EQUATION 
IX THE OTHER EQUATION 

Section 97. This method of elimination will be illus- 
trated by working the same problem which we used in 
the previous section. 



Solving Equations with Two Unknowns 201 

Illustrative example. Find the value of x and of y in 

the following equations : 

f*+ $(=10, (1) 

l*-3y = -6. (2) 

Solving equation (1) for x, we get 

x = 10 - y. (3) 

Substituting 10 — z/ for x in (2) gives 

10-y-3y = -6, (4) 

or _4y=-16, (5) 

or y = 4. (6) 

Substituting 4 for y in (1) or (2) gives 

x = 6. (7) 



Here, as when we eliminate one unknown by " compari- 
son," our real aim is to get an equation which contains only 
one unknown. We found from equation (1) that x = 10 — y. 
This value of x must be true for both equations. (Recall 
that x is the same for both equations, or for both lines, at 
their point of intersection.} For this reason we may sub- 
stitute 10— y in place of x in the second equation. This 
gives an equation in one unknown ; namely, y. 

The same results could have been obtained by finding 
the value of y, instead of the value of x, from one of the 
equations and substituting it in the other equation. For 
example : 

Second illustration of the method of eliminating by sub- 
stitution. 

Jx+ y = io. (l) 

U-3z/=-6. (2) 

From equation (1), y = 10 — x. (3) 
Substituting 10 — x for y in (2) 

x _3(10_x) = - 6, (4) 

or x - 30 + 3 x = - 6, (5) 

or 4x = 24, (6) 

or x = 6, (7) 
and y — 4, as before. 



202 Fundamentals of High School Mathematics 

EXERCISE 92 

Solve by the method of substitution and check each 

result : 

fr-2j=10 \y = x 

' \ Zx + 2y= 6 ' }3.r + 4j/=7 

Lr-3j=-l f^-+j = 8 

)4.r-j = -15 \2^— j/ = 10 



2. 



7. The sum of two numbers is 102 ; the greater 
exceeds the smaller by 6. Find the numbers. 

8. The difference between two numbers is 14, and 
their sum is 66. Find the numbers. 

9. 12 coins, nickels and dimes, amount to $1.05. 
Find the number of each kind of coin. 

10. The perimeter of a rectangle is 158 inches ; the 
length is 4 feet more than twice the width. 
Find the dimensions of the rectangle. 

11. Bacon costs 10 cents per pound more than 
steak. Find the cost per pound of each if 
4 pounds of bacon and 7 pounds of steak to- 
gether cost $3.48. 

12. A part of $4000 is invested at 4% and the re- 
mainder at 5%. The annual income on both 
investments is $185. Find the amount of each 
investment. 

13. The quotient of two numbers is 2, and the 
larger exceeds the smaller by 7. Find the 
numbers. 

14. Oranges cost 20 cents per dozen more than 
apples. A customer bought 10 dozen oranges 



Solving Equations with Two Unknowns 203 

and 4 dozen apples and received 20 cents in 
change from a 5-dollar bill. Find the price per 
dozen of each. 



15. 



16. 






17. 



lf/ = 



<7 — 4 

y 2 



2£_3j/_7 
X 4 ~~3 



18. 









III. ELIMINATION OF ONE UNKNOWN BY ADDING OR BY 
SUBTRACTING THE EQUATIONS 

Section 98. A great many equations in two unknowns 
can be most easily solved by this method. The following 
example illustrates it. 

Illustrative example. Find the value of x and y in these 

equations : 

(2x-y = 5, (1) 

{ x + y = \Z. (2) 
Adding equation (1) and equation (2) gives 

3jc = 18, (3) 

or x = 6. (4) 

Substituting 6 for x in (1) and (2) gives 

y = 1. (5) 
Check: Substituting 6 for x and 7 for y in (1) and (2) gives 

12-7 = 5. ^ (6) 

6 + 7 = 13. (7) 



It happens in this example that one of the unknowns, 
y t is eliminated by adding the corresponding members of 
the given equations. In many examples it is possible 
to eliminate one of the unknowns by subtracting the mem- 
bers of one equation from the corresponding members of 



204 Fundamentals of High School Mathematics 

the other. In many other examples, however, it is im- 
possible to eliminate one of the unknowns directly, either 
by adding or by subtracting the members of the two 
equations. For illustration, take this set of equations : 

Illustrative example. 

f*-2y = 8, (i) 

sx + y = e, (2) 

If we add the corresponding members of the two equa- 
tions, we get the equation 3x — y = 14. But this does not 
eliminate either of the unknowns. In the same way, if 
we subtract (2) from (1), we get the equation — x — 3 y = 2. 
Again, this does not eliminate either one of the unknowns. 
This shows that addition or subtraction of the members 
to the equations will not eliminate one of the unknowns 
unless one of them, the x or the y> has the same coefficient 
in both equations. 

Now let us make y in the second equation have the same coeffi- 
cient as y in the first equation. To do so, the second equation 
must be multiplied through by 2. This gives 

f4x + 2y = 12, (3) 

I *-2y = 8. (1) 

Now, by adding (3) and (1), we get rid of y, obtaining : 

5 x = 20, (4) 

or x = 4. (5) 

Substituting in (1) or (2), y = - 2. (6) 

Check : 16 - 4 = 12. (7) 

4 + 4=8. (8) 



An important question naturally arises here : When do 
we eliminate by addition and when by subtraction ? This 
can be answered by referring to an example. 

\x+y=U, (1) 

2x+y = 4. (2) 



Solving Equations with Two Unknowns 205 

Would either x or y be eliminated by adding the corre- 
sponding members of these equations ? Certainly not, for 
that would give Sx + 2y = 15. Now, would either x or y 
be eliminated by subtracting the members of one equation 
from those of the other ? Yes, for we should have — x = 7. 
However, if the second equation (2) above had been 
1x — y — 4, then we should eliminate jj/ by adding {1)2li\& (2). 
From these examples we come to the following conclu- 
sions about eliminating one of the variables : 

I. If the variable we wish to eliminate has the 
same sign in both equations, then it is elim- 
inated by subtracting the members of one equa- 
tion from the members of the other equation. 
II. If the variable we wish to eliminate has dif- 
ferent signs in the two equations, it is elim- 
inated by adding the corresponding members 
of the equations. 
III. No variable can be eliminated either by addi- 
tion or by subtraction unless it has the same co- 
efficient in both equations. If it does not have 
the same coefficient in both equations, then we 
must multiply one, or both, of the equations by 
such a number, or numbers, as will make that 
variable have the same coefficient. Thus, to 
eliminate x in the following equations : 

J2*- y = 8, (l) 

\3x + 4z/ = 23. (2) 

It is necessary to multiply (1) by 3 and to multiply 
(2) by 2. This gives the following equations : 

f6x-3y = 34, (3) 

\6x + 8y = 46. (4) 

Now the variable x can be eliminated by subtracting 
(4) from (3), which gives 

-Hi/ --22, 
or y = 2, 



2o6 Fundamentals of High School Mathematics 

EXERCISE 93 

ELIMINATION BY ADDITION OR SUBTRACTION 

Solve and check each of the following : 

lx+y = b [2>a-b = -2 

• \x-y = 2 [4tf + £ = -12 

l2x+3y = U I2x-j = ll 

2 - \3.r-3j = l 6- \x- 3^ = 13 
I2r+s = 9 i3b + 2c = 5 

3 - \r-s = 12 T \2& + c = 3 

fr-j/ = -8 (4jr-3y = 8 

4 I 8 I 

' \ x +y = -4: * \x-Ty = 2 

9. Find two numbers whose sum is 100 and whose 
difference is 18. 

10. In an election of 642 votes an amendment was 
carried by a majority of 60 votes. How many 
voted yes and how many no ? 

11. The admission to a school play was 25 cents for 
adults and 15 cents for children. The proceeds 
from 267 tickets were $ 50.05. How many tickets 
of each kind were sold ? 

12. A purse containing 18 coins, dimes and half 
dollars, amounts to $6.20. Find the number 
of each denomination. 

\/>-g = 8 \x+2y = ll 

13. V?, .„ 16. 



14. 



15. 



3/ + 4? = 10 }5;r-3 = 3j/ 

|- x +y = 3 J 3 x — y = — 2 

2x + 3y = U 1? ' I*- 3.7 = 10 
\2x + 5y = 12 j2a-8 = -b 

y-? jX =-l 1Sm 13^ + 4^ = 7 



19 

1 x 



Solving Equations with Two Unknowns 207 
l*+8 = -4, on ff*+f>=12 



^ 20 < * s-' 



In the remaining examples of this exercise, use any 
method of elimination. 

21. i^ + 4 . 25. J*=- 2 ^- 



2;r+j/=16 ' 1 7 = * +14.5 

>-j=10 26 * 12^-3^ = 7 

^=j/ + 2 * im \y + x =-lb 

4^ = 3j/ + 3 \2x+2y = 

24. { r / 28. 



5y = 6x \x =y + 12 

SUMMARY 

This chapter has taught three methods of solving a pair 
of equations which contain two unknowns : 

1. Elimination of one of the unknowns by substitution ; 

2. Elimination of one of the unknowns by comparison; 

3. Elimination of one of the unknowns by addition or 
subtraction. 

REVIEW EXERCISE 94 

1. If one tablet costs b dollars, what will x tablets 
cost? 

2. If a books cost b dollars, what will one book 
cost ? c books ? 

3. What is the perimeter of a rectangle whose 
width is a and whose length is b ? What is its 
area ? 

4. What is the width of a rectangle whose perime- 
ter is/ and whose length is xl 



208 Fundamentals of High School Mathematics 

5. The area of a triangle is k. Its base is b. 
What is its altitude ? 

6. The sum of two numbers is s. If one is d y 
what is the other ? 

Solve each of the following pairs of equations by any 
method of elimination : 

r — w 2 v = X r k ~ -. 

+ 4=7 



3,r+2; 



7 f5* 



6 4 

T""8 = 8 



8 [4^-^ = 10 

10. A collection box contained 63 coins, nickels and 
quarters. How many of each kind were there 
if the total amount was $ 8.35 ? 

11. Express algebraically that the weight, W, of 
the water in a tank varies as the volume, V, of 
the water. If 6 cu. ft. weigh 374.4 lb., how 
much will 11 cu. ft. weigh ? 

12. In weighing with a spring balance scale, the 
principle is applied that the amount of stretch, 
s, of the spring varies as the weight, W, which 
is suspended to the scales. Express this more 
briefly. If a 20-pound weight produces a stretch 
of | inch, what weight will produce a stretch of 
2 inches ? 

13. What is the best method of eliminating one of 
the unknowns in an equation ? 

14. The volume, V, of a sphere varies as the cube 
of its radius, r. Express this more briefly. If 
an orange with a radius of 2 inches is worth 10 
cents, what could you afford to pay for an 
orange with a 4-inch radius ? 



Solving Equations with Two Unknowns 209 



15. 






The table below shows how much money (to 
the nearest dollar) you would have at the end 
of a certain number of years if you saved 10 
cents a day and deposited it in a bank which 
pays 3 % interest. 



At the end 
of (years) 


1 


2 


3 


5 


8 


10 


14 


17 


20 


total amt. 
saved is 


37 


75 


H5 


197 


330 


425 


635 


809 


999 



16. 



17. 



Represent this graphically, measuring the time 
on the horizontal axis. 

Estimate the total savings at the end of -i yr. ; 
6 yr. ; 25 yr. 

Is it ever possible to get from a graph informa- 
tion which could not be obtained from the 
table ? Illustrate. 

The distance from the base to the top of a hill, 
up a uniform incline of 40°, is 800 ft. What is 
the altitude of the top above the base ? 



CHAPTER XIII 

HOW TO FIND PRODUCTS AND FACTORS 

Section 99. Why you should be able to find products. 

Suppose you wanted to find the area of a rectangle whose 



X 

(SI 



3X 



Area=6-X 2 



Area= 



-3Xt5 

Fig. 104 



cm 






dimensions are 8;r + 5 and 2x. To do so, it would be 
necessary to multiply 3x+ 5 by 2x, or to find the product 
of these two algebraic expressions. One way to do this 
is to divide the rectangle into smaller rectangles, as indi- 
cated in Fig. 104. This gives two rectangles, the dimen- 
sions of one being 2x by 3x, and of the other 2x by 5. 
From what you already have learned about multiplication 
you can see that the areas of these are 6x 2 and 10. r, be- 




How to Find Products and Factors 211 

cause Sx times 2x is 6x 2 and 5 times 2x is 10 x. Thus 
the area of the original rectangle is 6x 2 + 10 x. Similarly 
the area of the rectangle in Fig. 105 is what ? What 
would its area be if the dimensions were Qa + 4 and 5 a ? 

These illustrations are given to make clear the need of 
learning how to find products. Other illustrations might 
have been taken. For example, what is the cost of 
15 b + 3 articles at 4 b cents each ? How much could you 
earn in 6 y + 4 days at Sy dollars per day ? 

A NEW WAY OF INDICATING MULTIPLICATION 

Section 100. As you progress through your mathe- 
matics, you will find that it is a language which tells more 
in fewer words or symbols than any other language. For 
example, instead of writing "find the product of 3x+5 
and 2 x" it has been agreed to express this by means of the 
parentheses, ( ). Thus, 2.r(3;r+5) means " to find the 
product ofSx+o and 2x," or, "to multiply 3 x + 5 by 
2x. It is important to note that there is no sign between 
the 2x and the expression in the parentheses. Similarly, 
to state algebraically the problem in the second illustra- 
tion, Fig. 105, you would write 4x(7x + 3), putting 
no sign between the 4:X and the parentheses. Thus, 
5 b(3 b + 7) means to multiply each of the numbers in the 
parentheses by 5 b. 

ORAL EXERCISE 95 
PRACTICE IN FINDING PRODUCTS 

In the following examples, multiply each term within the 
parentheses by the number which immediately precedes 
the parenthesis, or, remove parentheses. 



212 Fundamentals of High School Mathematics 
Illustrative example. 3 *(5 x 1 + 7 x + 8) = 15 x 3 + 21 x 1 + 24 x. 



1. 4 j'(3j' + 9) 10. 8 j/ (7 -8) 

2. (i/;(i>/; + l) ii. 3r(r+3) 

3. 7 £:(3 + 5 c) 12. lb{l-b) 

4. f), r (.r-4) 13. 8(/; 2 -8£ + 12) 

5. 9(2** + 7* -4) 14. 6(2*-3£ + <r) 

6. 4a(a 2 +3a + 7) 15. ab{a + b + l) 

7. 6j(3j> 2 - 5j/ + 2) 16. ^/ 2 (^r + j/ + ^) 

8. 1(2^ + 3) 17. -1(4 -5 y) 

9. 5/(6-/) 18. -3^r(6^-4) 

19. What algebraic expression will represent the 
area of a rectangle whose length is 10 inches 
more than its width ? 

20. What algebraic expression will represent the 
total daily earning of 4 men and 7 boys, if each 
man earns 1 2 per day more than each boy? 

21. -6(2*- 7) 24. (16/ -7) 

22. -(10-*) 25. (a + b) 

23. _4j/ 2 (2j/-3) 26. -(b-c) 

In this exercise you have learned how parentheses are 
used to indicate that each of the terms in an expression 
must be multiplied by another number. 

Section 101. More difficult multiplication. Most prod- 
ucts which you will need to find are more difficult than 
those of the preceding exercise. For example : How 
many square feet of floor area in a dining room whose 
dimensions are 4^r + 3 ft. and 5x + 4 ft? To find the 



How to Find Products and Factors 



213 



T 



X 



5 X 



-M--4- 



Area = 15 x 



Area = 20 X 2 



Area=12 



Area = 
16 x 



-Sx+4 

Fig. 106 



product of these factors requires something you have not 
yet learned. You know how to multiply bx + 4 by 4 x or 
4„r + 3 by 5x, but you have not learned how to multiply 
such expressions as 4^ + 3 by bx + 4. The drawing 
shows one way to do this ; namely, the geometrical method 
of dividing the entire area into smaller rectangles, the 
area of each of which you can find. This method gives 
four rectangles whose areas we can find. Thus, we get 
four rectangles whose areas are 20 ;r 2 , 16.r, 15 x y and 12, or, 
collecting terms, an entire area of 20 x 2 + 31 x + 12. 

Another method of finding the product of 4,r + 3 and 
5x + 4 (which is generally written as (4„r + 3)(5x + 4)) 
makes no reference to rectangles. It is very much like 
the method of multiplication used in arithmetic. To illus- 
trate, this same example could be solved as follows : 



2i4 Fundamentals of High School Mathematics 
First illustrative example. 

(4jc + 3)(5x + 4). 

4jc + 3 
5x-f-4 



20x 2 + 15* 

+ 16s -f 12 
20jc 2 + 31jc + 12 



The 20 ;r 2 + 15 .r is the result of multiplying 
4.r + 3 by bx, and the 16^ + 12 is the result of 
multiplying \.x + 3 by 4. This latter method is 
much more generally used than the geometrical 
method. Let us take another illustration. 
Second illustrative example. 

(7 a + 3) (4 a -5). 

7a + 3 

4q-5 
28a2 + 12 a 

- 35 a - 15 
28 a3 - 23 a - 15 



Note that 7 a + 3 was first multiplied by 4 #, giving 
28tf 2 +12<*. Then 7 * + 3 was multiplied by -5, 
giving — 35 a — 15. How was the final product ob- 
tained ? 

Which of these two methods, do you think, should 
be learned ? 



How to Find Products and Factors 



215 



EXERCISE 96 
PRACTICE IN FINDING PRODUCTS 

Illustrative example. 

(2a: 2 + 3x+4)(3x- 7). 
2x 2 +3x +4 

3s-7 

6x 3 + 9x 2 + 12* 

-14s 2 - 21s -28 
6x 3 - 5x 2 - 9x-28 



1. (3^ + 2)(4^ + 5) 11. 

2. (7£ + 5)(> + 6) 12. 

3. ( 7 + 4)(5j/ + 3) 13. 

4. (/-5)(2/+8) 14. 

5. (8j/ + 5)(7j/ + 3) 15. 

6. (6^ + 4)(4^ + 6) 16. 

7. (4*-3)(4a-8) 17. 

8. (*+3)<>r-9) 18. 

9. (3^-4)(3^-9) 19. 
10. (6/-7)(3/-9) 20. 

21. 



7 + 6)<j/ + 6) 
3£ 2 +l)(2£2 + 5) 

4^ 3 + 3)(3^ + 10) 
5a+3)(5a -3) 
7^ + 5)(7r — 5) 
# + 7)(<z — 7) 
3^-2)(3^-2) 
x ?+2x+l)(x^-3) 
_T 2 + 6j/ + 9)(j/ + 3) 
^ 3 + 5)(^ 3 -5) 



If you should multiply x + 7 by .r + 10, what 
would be the first term of your product ? What 
would be the last term ? 
22. Can you tell, at a glance, the first and last 
terms of the product which you would obtain 
by multiplying 2x + 1 by 5 x + 4 ? 

A SHORTER METHOD OF MULTIPLYING 

Section 102. There is a much shorter method of finding 
products like those in Exercise 96. Mastery of this short 



216 Fundamentals of High School Mathematics 

cut will not only save a great deal of time, but it will help 
you in the later work of this chapter. For an illustration, 
take the example 

(3£ + 5)(2£ + 7). 

You have no difficulty in seeing that the first term of the 
product is 6 b 2 {i.e. 3 /; x 2 b) and that the last term is 
+ 35 {i.e. 5 x 7). So if there were some method by which 
you could tell the middle term, you could write the product 
at once, without using the longer method of placing one 
factor under the other and multiplying in the regular way. 
To make the new method clear, it is necessary to refer 
again to the regular way. Let us illustrate with the 
example : 

Illustrative example. 

(3ft + 5)(2ft+7). 
By the old method : 

6 ft 2 + 10 ft 

+ 21 ft + 35 
6 ft 2 + 31 ft + 35 

The arrows show the cross-muViplications or cross-products 
that make up the middle term. One " cross-product " is 2 ft times 
+ 5, or + 10 ft, and the other cross-product is + 7 times + 3 ft, 
or + 21 ft. Combining the cross-products, we get the middle 
term, + 31 ft. 
By the shorter method we get 

+ 21 ft 
(3ft+5)(2ft+~7) = 6 ft 2 + 31 ft + 35. 

+ 10 ft 



The curved lines indicate the cross-multiplication, or 
cross-products, which must be combined to give the middle 
term, + 21 £ and +10£, giving +S16. Thus, in this 



How to Find Products and Factors 217 

new method, it is assumed that you can tell, at a glance, 
the first and last terms of the product. Then you can 
get the middle term by finding the sum of the cross-prod- 
ucts. The curved lines are drawn to help you see the 
cross-products. 

Second illustrative example. 

(4& + 3) (7 &- 8). 
The Long Method The Short Method 

-32& 
(4 b +3) (7 b -8) = 28 b°— 11 b - 24 
+ 21b 
24 




28 fr 2 - 11 b - 24 



It is important to recognize that the curved lines indicate 
cross-products in just the same way that the arrows in the 
long method refer to cross-products. The new method, 
which from now on we shall call the cross-product 
method, enables you to do mentally in much less time 
what was written down by the old method. To give you 
practice in this important method of finding the product of 
two factors the following exercise has been included. 

EXERCISE 97 

Find the products of the following factors by the cross- 
product method : 

1. (* + 2)(jr + 5) 6. (5£ + 3)(0 + l) 

2. (2j/ + 4)(3j+5) 7. (9* + 2)(2;r+l) 

3. (£+6)(2£+7) 8. (4^ + 3X7^+10) 

4. (.r+4)(3.r + 5) 9. (5j + 3)(5j-3) 

5. (/-8)(*-3) 10. + 9)0 + 9) 



218 Fundamentals of High School Mathematics 



u. 

12. 
13. 
14. 
15. 
16. 
17. 
18. 
19. 
20. 
21. 
22. 
23. 
24. 
25. 
26. 
27. 
28. 
29. 
30. 
31. 



* + 2)(f— 5) 32. 

rt/> + <'>)(,?£ + 3) 33. 

.r 2 +3)(.r 2 + <;) 34. 

rt<V+8)(rt/;c-10) 35. 

■2.V+ 3j')(2.r + 3j) 36. 

a + £)(a + £) 37. 

c + d){c + d) 38. 

f+s)(f+s) 39. 

.r 4- 5)(.r - 5) 40. 

,r + 10)(,r-10) 41. 

J + 4)(j-4) 42. 

3/ + 7)(3/-7) 43. 

4rt4-3)(4rt + 3) 44. 

7£ + 9j)(7£ + 9j/) 45. 

> + 3)(^ + 5) 46. 

x - 8)(x + 8) 47. 

c + 9)(c + 9) 48. 

^-4)(7-4) 49. 

3« + 2)(2a + 4) 50. 

4 / -2)(3 7 + 5) 51. 

2^ + 4)(5^--9) 52. 



y + 4)(rf+7) 
c- 9)^ + 9) 

7 + 7)( 7 + 7) 
a - 2)(a - 2) 
5a + l)(a-2) 
2.r+l)(x-2) 
4j + 3)(3j-9) 
/2 + 3)(7 2 + 5) 
/»-9)(/ + 2) 

r 2_3)( r 2_ 5) 

y + 3)(/ + 3) 
^ 3 -ll)(^-ll) 
* 2 + 3)0 2 + 3) 
4j+10)(5j-8) 
a^ + 2r)(rt^ + 3r) 
<6abc — d)(2abc + d) 
lx 2 t+y)(2x 2 t-5j/) 
9.r 2 +3j)(9;r 2 + 3j/) 
8«-4/)(8a-7^) 
/+ *)(/+*) 



How to Find Products and Factors 



219 



Area= 
X 2 + 12X+36 



53. Illustrative example. What expression will repre- 
sent the area of a square if each side is x 4- 6 inches ? 

Solution : Evidently the area is 
(x + 6)(x + 6), or x°- + 12*4-36. 
This is usually written, however, 
not as (jc + 6)(x + 6) but as 
(x + 6) 2 . The exponent, 2. shows 
that x 4- 6 is used twice as a fac- 
tor. Thus, (3 x + 5) (3 x 4- 5) is 
usually written as (3 x 4- 5) 2 . In 
the same way. (3 x 4- 5) (3 x 4-5) 
(3x+5) would be written as 
(3x4-5)3. 



X+6 - 

Fig. 107 



(0 



54. 


+ 5) 2 


55. 


(3 £+5) 2 


56. 


(2y + 7f 


57. 


(4 a - 3) 2 


58. 


(x+yf 


59. 


{x-yf 



66. 







60. 


(a + 


*y> 






61. 


(a- 


bf 






62. 


(/4 


■sf 






63. 


(/- 


sf 






64. 


(2* 


+ 3 y) 






65. 


(4/; 


-2)2 


(a 


+*)(« 


+ d) 





Section 103. Further practice in translating from alge- 
braic symbols into word statements. You have had some 
practice in translating from algebraic statements into 
word statements. For example, you translated the alge- 
braic expression " a + b " into the word statement "the 
sum of two numbers," and the expression "xy" into "the 
product of two numbers." It is important to be able to 
translate into word statements some of the examples which 
you did in the last exercise. 



220 Fundamentals of High School Mathematics 

EXERCISE 98 

Write out the word statement which means the same 
thing as each of the following algebraic expressions : 

1. a-b 7. (a + bf 

2. a* 8. (a-bf 

3. a* 9. ( a + b)(a-b) 

4. (2j>f 10. r* + s s 

5. c + d 11. (f+sf=f* + 2fs + s 2 

6. a 2 + b 9 12. (f-sf=f*-2fs + s* 
Section 104. How to solve equations which involve prod- 
ucts. The solution of a great many equations depends 
upon your being able to find products like those you have 
just been finding. To illustrate, consider the problem : 

Illustrative example. The length of a rectangle is 6 inches 
more than, and the width is 2 inches less than, the sides of a 
square; the area of the rectangle exceeds the area of the square 
by 20 square inches. What are the dimensions of each ? 

Solution : Translating into algebra, we have the equation : 

(s+ 6)(s-2)=s 2 + 20. 
Multiplying, or removing parentheses, gives 

s 2 + 4 s — 12 = s 2 + 20. 
Subtracting s 2 from each side gives 

4 s - 12 = 20. 
.-. s = 8. 
Thus, the sides of the rectangle are 14 and 8. 
Check this result. 






EXERCISE 99 
PRACTICE IN SOLVING EQUATIONS WHICH INVOLVE PRODUCTS 

Solve and check each of the following equations : 

1. ( jr + 6) 2 = ;r 3 + 96 

2. (j' + 8)0' + 6)-y ! = 63 

3. (2£ + 8)(£ + 4)=(£ + l)(2£ + 10) 






How to Find Products and Factors 221 

4. One number is 5 larger than another ; the 
square of the larger exceeds the square of the 
smaller by 55. Find each number. 

5. The length of a rectangle is 8 inches more 
than, and its width is 3 inches less than, the 
sides of a square ; the area of the rectangle ex- 
ceeds the area of the square by 26 square inches. 
Find the dimensions of the rectangle. 

6. 0' + 5)0/-5) = (j/-6)( 7 + 2) 

7. 2(x- 8)- 3(,r-4)=- ox 

9. (^ + 3)2_(^ r _ 1)2 = 40 

10. ( 7 + 5) 2 -(j + 4) 2 = -l 

11. (x- 8f=(x- 12) 2 

12. 2(/+3) 2 =2(/-8) 2 

13. 3(x+6)(x + ±) = (Sx+l)(x+9) 

II. HOW TO FIND THE FACTORS OF AN ALGEBRAIC 
EXPRESSION 

A. FINDING THE COMMON FACTOR 

Section 105, Meaning of the word FACTOR. If you 

know that the area of a rectangle is 24 square inches, 
what might be its dimensions ? You readily see here that 
the dimensions might be 4 inches and 6 inches ; or 3 inches 
and 8 inches ; or 12 inches and 2 inches, because the 
product of 4 and 6, or of 3 and 8, or of 12 and 2, is in each 
case 24. This process of finding the members which, when 
multiplied together, give another is called FACTORING. 
The numbers you find are called the FACTORS. Thus, 
4 and 6 are factors of 24 ; also 3 and 8, or 12 and 2. 



222 Fundamentals of High School Mathematics 




Fig. 108 



The same reasoning is used in algebra as in arithmetic. 
For example, suppose the area of a rectangle is 5^+35 
square units. If the width is 5 units, what must the length 
be ? Similarly, if the area is 4:x 2 + 28 ;r and the width is 
4;r, what is the length ? 

Section 106. What is a common factor ? Now let us 
take a more difficult illustration. In the previous two 
cases, you knew both the area and one dimension, or the 
product and one of its factors. But in most factoring prob- 
lems you do not know any of the factors. For example, 
what are the dimensions of the rectangle whose area is 
7 x + 21, or, in other words, what are the factors of 
Ix -+-21? A study of the 7x and 21 shows that 7 is 
a factor of each, or is a COMMON FACTOR of both 
terms. What, then, must 7 be multiplied by to give 
1 x + 21, or, what must be the length of this rectangle if 
its width is 7 ? Evidently, it must be x + 3. This prob- 
lem should be written 

7^ + 21 = 7(> + 3). 

A second illustration should make clear what is meant by 
factoring in algebra. 

Find the factors of ax + ay + aw, or find the dimensions 
of a rectangle if its area is ax + ay + aw. 



How to Find Products and Factors 223 

By observing each term, we see that a is a factor common 
to each. 

Dividing each term of the expression by a gives the other 
factor, x+y+w. Hence, ax + ay + aw = a(x+y + w), 
and a is one factor and x +y + w is the other one. 

These illustrations are intended to make clear how to 
factor an expression which contains a common factor. 



EXERCISE 100 

Factor each of the following expressions : 



1. 


3^+12 


15. 


5x 2 —by 2 


2. 


5^-20 


16. 


4, a 2 + 8ab+4,P 


3. 


ax + bx 


17. 


X 2 + X 


4. 


la -21 


18. 


a 2 + 20 cP 


5. 


llx + U 


19. 


x 2 + 5x s 


6. 


6a + 9 


20. 


a + ab + a 2 


7. 


5 + 15 a 


21. 


x 2 -bx± 


8. 


9 + 6^ 2 


22. 


4a*-12aZ 


9. 


4^ 2 + 12 


23. 


5 x 2 y — 5 xy 2 


10. 


Sa + 12b 


24. 


7b*-21 b 2 


11. 


ab + ac + ah 


25. 


12ab +6# 


12. 


5 + 10 a + 15 a 2 


26. 


x 2 + 2 xy 


13. 


2x 2 -±xy + 2y 2 


27. 


2- 20x 


14. 


2-8d 2 


28. 


64^ 2 -21^ 




29. 6 ax 2 — 12 ax s 



Can you check the examples in this exercise? Check 
this one : 

5* 2 -15;r 4 = 5;r 2 (l--3;r). 



j 24 Fundamentals of High School Mathematics 



B. THE (ROSS-PRODUCT METHOD OF FACTORING 

Section 107. In the previous section all the expressions 
which you factored had a common factor. But most ex- 
pressions which you will need to factor are much more 
difficult than those ; and they do not always contain a 
common factor. 

Illustrative example. Factor 2 x 2 + 5 x + 3, or find the di- 
mensions of a rectangle having this area. 

From what you learned about products, you can see 
that this expression was very likely made by multiplying 
two factors together. Also, you can see that the first 
terms of the two factors must be 2x and x. To help you 
to get the correct result, always write the blank form of 
the two parentheses first, thus: ( )( ). Then as 
you determine each term of each factor, you can write it 
in the appropriate place. Later you will doubtless be 
able to do all the work in your head and not have to 
write out the steps. Second, therefore, write the first 
terms in the blank form, thus : 

2jc 2 +5jc+ 3=(2x )(x ). 

Thirdy you have to find the second terms of each factor. 
From your previous work in finding products, you know 
that the last term of the expression, + 3, was obtained by 
multiplying together the second terms of the two factors. 
Then the second terms must be such that their product is 
+ 3. Obviously, they are 1 and 3 or 3 and 1. To tell 
whether the 3 or the 1 belongs in the first factor we have 
to try it, and test or check to see if the middle term will 
be correct (+ bx). Trying this out, we have : 

ex 



2x 2 + 5jc + 3 =(2jc + l)(x + 3). 
x 



Haw to Find Products and Factors 225 

Checking, — that is, multiplying the two factors together, 
— we see that this does not give the correct middle term, 
for + 6;r and + x are not ox. But, we might interchange 
the 1 and 3. Trying this, we get 

2x 



2x' 2 + bx + 3 = (2x + S)(x + 1). 
3jc 

Multiplying these together shows that our factors are 
correct, for their product gives the original expression, 
2. r 2 + ox + 3. 

Let us try another example : 

Second illustrative example. Factor 5 x- - 36 x + 7. 
First write the blank form thus : ( )( ). 

Next we can tell at once that the first terms of our 
factors are ox and x, giving 

5x*-36x + 7 = (5x )( X ). 

Examining the last term of the expression, + 7, we 
see that the second terms of the required factors must be 
1 and 7 or 7 and 1. Trying out the 1 and 7 gives 

5x*-36x + 7= (5x+l)(x + 7). 

But the check shows that the sum of the cross-products 
is + 36 x, whereas it should be — 36 jr. This can be cor- 
rected by changing the sign of the second terms to — 1 

and - 7, giving 

(5x-l)(x-7). 



The result of checking shows these to be the correct 
factors. 



220 Fundamentals of High School Mathematics 

Section 108. The sum of the cross-products must equal 
the middle term. These two explanations have been given 
to show the importance of getting, as factors, expressions 
such that the sum of the cross-products will give the middle 
term of the expression to be factored. It is assumed that 
you can tell, at a glance, what the first terms might be, and 
what the second terms might be by looking at the first and 
last terms of the expressions which you want to factor. 

For example, in factoring 

6;r 2 + 13.r+6, 

the first terms might be Sx and 2x } or 6x and x\ the 
second terms might be 3 and 2, 2 and 3, or 6 and 1. But 
since the product of the factors must give the original 
expression, we can tell by trying these various possible 
combinations that the factors are 

±x 



(3* + 2)(2;r + 3), 

~Yx 

No other arrangement of numbers will give the correct 
middle term, + 13 jr. 

From this explanation you should be able to factor the 
expressions in Exercise 101. Don't be discouraged if you 
have to try more than once before you succeed. Difficult 
tasks often require many trials. 



How to Find Products and Factors 227 

EXERCISE 101 
FACTORING BY THE CROSS-PRODUCT METHOD 

Check each example carefully. 

(The parentheses are written here to suggest to you 
how to begin.) 

1. ** + 5*+6 = ( )( ) 

2. 7 2 + 10j/-21 = ( )( ) 

3. 2x 2 + 7x + 5 = (2x+?)(x+?) 

4. c 2 + 6c + 9 = ( )( ) 

5. zZ-8x+12=( )( ) 



6. 


5/ + I67 + 2 = (5. 


r+?)(j+?) 


7. 


a 2 + 12 a + 36 = 


20. 


21 b 2 - b - 2 = 


8. 


5x* + 8x + 3 = 


21. 


5.r 2 -6^ + l 


9. 


x*-Zx-U = 


22. 


3^2 + 4r + i 


10. 


m 2 — m — 20 = 


23. 


2/ -j -28 


11. 


2 b 2 + 13 b + 15 = 


24. 


2 a 2 + 7 a + 3 


12. 


3^ 2 -13^-t-4 = 


25. 


^2_n r + 21 


13. 


/*--5/-40 = 


26. 


y _ i 0j/ + 20 


14. 


IO7 2 + 13 jp - 3 = 


27. 


^ 2 + 6 a + 9 


15. 


4^ + 20^+25 = 


28. 


y 2 -8y + 16 


16. 


a 2 + # - 72 = 


29. 


y + 10 


17. 


y - 16 = 


30. 


^2 + ^ _ 30 


18. 


£ 2 - 25 = 


31. 


3^r 2 + 8^ + 5 


19. 


15 c 2 _ 31 , + 14 = 


32. 


2^2_ 5;r + 3 



Section 109. Importance of finding the prime factors. 

Any algebraic expression that cannot be factored is prime. 
For example, 3 x + 5 is prime, because there are no integral 
expressions which can be multiplied together to produce it. 



228 Fundamentals of High- School Mathematics 

But 9jt+6 is not prime because it can be obtained by 
multiplying 3 and x + 2. 

It is important that you should always find prime factors. 
To illustrate : 

First illustrative example. Factor 3 b 2 - 21 b + 36. 

By inspection, we see that 3 is a common factor. 
Removing it, we have 

S(b 2 - 7 6+ 12). 

Now. unless we remember that prime factors should be found, 
we are likely to leave the example in this incomplete form. 
The b 1 — 7 b + 12 can be factored further, however, giving 

(*-3)(*-4). 

Thus, the original example should be factored as follows : 
3 b 2 - 21 b + 36 = 3(62 - 7 b + 12) 
= 3(6-- 4) (b -3). 



Second illustrative example. Another illustration will 
make clear the importance of finding prime factors. Suppose we 
wish to factor 2 x 2 — 50. As in the previous example, we always 
first look for a common factor. This gives 

2(x*-25). 

Now, again, we are apt to leave the example in this incomplete 
form, not remembering to see if we can further factor x 2 — 25. 
We see, however, that we can. The complete solution is : 
2x 2 -50 = 2(x 2 -25) 

= 2(x + 5)(x-5). 



These explanations are given to help you keep in mind 
that in all factoring work there are two absolutely essen- 
tial steps ; namely, 

1. LOOK FOR A COMMON FACTOR. 

2. FIND PRIME FACTORS; I.E. FACTOR COMPLETELY. 



How to Find Products and Factors 229 



1. 


2^ 2 + 14^4-24 


2. 


5j/ 2 -45 


3. 


sfl - st- 20 s 


4. 


7^-14^-105 


5. 


3*3 + 12* + 45 


6. 


* 2 -6*+9 


7. 


6/ 2 -15 / 3 


8. 


2 - 128 / 2 


9. 


aP-a&- 72 a 


10. 


6* 2 + 13* + 6 


11. 


3 a 2 + a - 2 


12. 


2 ^ - 5 * + 3 


13. 


7 £ 2 - 17 £-12 


14. 


<f _ 12 £ - 28 


15. 


2* 2 -14* + 24 


16. 


^ 4 — 6 j/ 2 — 16 


17. 


49^+70^+25 


18. 


5a 2 - 80 


19. 


x z — X 1 


20. 


6* 2 -18* 3 


21. 


3y _ 3 y _ 36 


22. 


12* 2 + 37*-10 


23. 


10* + 25+* 2 


24. 


j 2 + 10 


25. 


* 2 + 36 



EXERCISE 102 
PRACTICE IN FACTORING COMPLETELY 

26. 2 a 2 -8 



27. 3£ 3 + 27 

28. 3^-12^-180 

29. 2 ^ + 10 x -168 

30. ;r 2 -.r-110 

31. 2j/ 2 -j/-1 

32. 6 a 2 -4a -2 

33. 3x 2 + ±x + l 

34. 20.r 2 + 70.i- + 60 

35. 2£ 2 -£-3 

36. 2rt 2 + 18 



37. 


5 y — 15 j 2 


38. 


a 2 -b 2 


39. 


3 J 2_ J _ 10 


40. 


2*2 + 3^_9 


41. 


6j/ 2 +7— 15 


42. 


8* 2 -2 


43. 


18 fl - 50 


44. 


9* 2 + 17*-2 


45. 


6^_^_ 12 


46. 


7 y _ 9 y _ 10 


47. 


2^-36* + 64 


48. 


j/ 2 — 3 j/ — 4 


49. 


/ 2 -16 


50. 


/ + 16 



230 Fundamentals of High School Mathematics 

SUMMARY 

This chapter has taught the following methods : 

1. A short method of multiplying which we call the 
" cross-product " method. 

2. How to find the factors of an algebraic expres- 
sion. 

REVIEW EXERCISE 103 

1. What is the area of a square formed by adding 
4 ft. to the sides of a square x ft. long ? 

2. What does (x— 4)(;r -f 6) represent, if x repre- 
sents the side of a square ? 

3. A rectangular field by rods long has a perimeter 
of 24 j rods. What expression will represent 
the area of the field in square rods ? in acres ? 

4. If the quotient is represented by q, the divisor 
by d y and the remainder by r, what will repre- 
sent the dividend ? 

5. If a park is w rods wide and 1 rod long, how 
many miles would you walk in going around it 
n times ? 

6. How do you divide a product of several factors 
by a number ? For example, in dividing 12-3-6 
by 2, would you divide each factor by 2 ? 

7. How do you multiply a product of several fac- 
tors by a number ? Give an illustration. 

8. The product of four factors is 60. Three of 
them are 2, 3, and 5. Find the fourth factor. 

9. How much do you increase the area of a square 
whose side is x, if you increase its side 4 units ? 



How to Find Products and Factors 231 

10. Translate into words : (/+ s) 2 =/ 2 + 2fs + s 2 . 

11. Solve for x y explaining each step : 

-4^+6 = 2^-18. 

12. In what way is factoring like division ? How is 
it like multiplication ? 

13. If an automobile uses 8| gallons of gas in going 
120 miles, how many gallons will it use in going 
250 miles ? 



14. Solve 1 



J 6 y -j=T + 4 y y 
I5;r + 8j/=l. 



15. Make up five examples for the class to factor, 
and then give them ^o the class to work. 

16. How many terms do you get when you square 
the sum of two members, e.g. (2x+3yf? 
when you square the difference of two numbers, 
e.g. (4 * - 3 bf ? 

17. Evaluate {a + bf if a = — 3 and b = + l. 

18. Does (a + bf = a 2 + b 2 ? Show by using 4 for a 
and 5 for b. 

19. If — 1 = ^2, what is the value of V x when V 2 

V 2 Pi 
= 40,^ = 8, and/ 2 = 12? 

20. A tree stands on a bluff on the opposite side of 
a river from the observer. Its foot is at an 
elevation of 45° and its top at 60°. Which has 
the greater height, the bluff or the tree ? What 
measurement would you have to make to find 
the height of the tree ? the width of the river ? 

21. Translate : a 2 - b 2 = (a + b)(a - b). 



CHAPTER XIV 

HOW TO SOLVE EQUATIONS OF THE SECOND DEGREE 

Section 110. What are quadratic equations ? In all 

the previous chapters you have solved equations of the 
first degree ; that is, equations in which the unknown (or 
unknowns} did not have exponents greater than 1. This 
chapter will show how to solve equations of the second de- 
gree, equations in which the unknown occurs to the second 
power. To illustrate, you will learn how to solve equations 
such as 

x 2 + 8.r=20. 

The fact that the unknown, x, in this equation occurs in 

the second power or second degree (as, x 2 ) leads us to 

i 

speak of the equation as a second-degree, or quadratic, 

EQUATION. 

Three ways of solving second-degree or quadratic equa- 
tions will be explained. These, in the order in which we 
shall discuss them, are : 

I. Solution by graphical representation. 
II. Solution by factoring. 
III. Solution by completing the square. 

Before we take up the first method, it will be necessary 
to study how to represent graphically algebraic expressions 
of the second degree which are not equations. 

A. HOW TO REPRESENT GRAPHICALLY AN EXPRESSION 
OF 'I'Hi; SECOND DEGREE IN ONE UNKNOWN 

Section 111. The value of an expression depends upon 
what value is assigned to x. Let us consider the second 
degree or quadratic expression 

2X2 



How to Solve Equations of the Second Degree 233 

Has this expression a definite numerical value ? What is 
it? 10? 18? If not, what? Evidently, we cannot tell 
what the value of the expression is unless we know what the 
value of x is. That is, the numerical value of the expres- 
sion depends upon what value is assigned to x. In other 
words, as x changes, the value of the expression x 2 — 8x 
+ 12 changes. Thus, we are dealing here with two vari- 
able quantities ; x itself is one variable, and the value of the 
expression jc 2 — 8 x + 12 is the other one. 

Section 112. (1) Tabulating values of the two variables. 
Thus, to represent an algebraic expression of the second 
degree graphically we first have to determine what numeri- 
cal value the expression has for various assigned values of 
x. That is, we will let x be 0, say, and we find by evalu- 
ating x* — 8x + 12 when x = 0, that the expression is 12. 
In the same way, when x = 1, the expression is 5 ; when 
x = 2, x*-8x+12 is 0, etc. 

We have already learned that to represent the way in 
which two related quantities change together, it is best to 
tabulate first the values of the two variables. For the 
present example we get the following table : 

Table 15 



If xis 





1 


2 


3 


4 


5 


6 


7 


-1 


-2 


then X-8X+12 is 


12 


5 





-3 


-4 


-3 





5 


21 


32 



etc. 



Section 113. (2) Plotting the graph of the two variables. 
With the table of values of the two variables [(1) the un- 
known, (2) the expression which contains it] once made, 
we can plot the points representing these pairs of values. 
Let us agree for convenience to plot the unknown, say x. 



234 Fundamentals of High School Mathematics 



is 

,2 



on the horizontal axis and the expression (in this case, 
x- — &x+ 12), on the vertical axis. Reading the pairs of 
values 0, 12; 1, 5 ; 2, ; etc., from the table, we obtain as 
the graphic representation of x 2 — §x+12 the curve of 
Fig. 109. 

Note that from the graph we can find the value of 
x 1 — Sx + 12 for any value of x represented on the .f-axis. 

For example, if x is 
Y 2 J, then ^ 2 -8^+12 

3| ; if x is 0, then 
8;r+12is 12, etc. 
Note also from the 
curve that x 1 — 8 x + 12 
is zero for two particu- 
lar values of x\ namely, 
x= + 2 and x= +6, 
i.e. tJie curve cuts the x- 
axis at + 2 and + 6. 

The graph shown in 
Fig. 109 is called a 
graph of an algebraic 
expression of the sec- 
ond degree, or a quad- 
ratic expression. 
Section 114. Graphs of second-degree equations are always 
curved lines. Turn back to Figs. 14, 16, 17, etc. What 
difference do you notice between each of the graphs in 
these figures and the one in Fig. 109 ? If you will turn to 
Fig. 110, you will check your conclusion. One new fact of 
importance is, therefore, that the graphs of expressions of 
the first degree are always straight lines, whereas graphs 
of expressions of the second degree are curved lilies. 



































































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f 


























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4d 


























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If 


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Fig. 109 



How to Solve Equations of the Second Degree 235 







































































































































































































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Fig. 110 



Graphs of such quadratic expressions are U-shaped and are 
commonly called parabolas. 

Figure 110 is a graphical representation of the quadratic 
expression x l — 2x ■— 35. By referring to it, you will be 
able to answer the following questions ; 



EXERCISE 104 

1. How many different values may the expression 
have ? 

2. What is the value of x when the value of the 
expression is zero ? 

3. For what value or values of x is the expression 
equal to — 10 ? 



236 Fundamentals of High School Mathematics 

4. What is the lowest or least value of the ex- 
pression ? 

5. I low were the points which fix the position of 
this curve located ? Are two points enough to 
determine the graph ? 

6. What is measured along the horizontal axis ? the 
vertical axis ? 

7. What variables are represented in this graph ? ■ 

EXERCISE 105 
PRACTICE IN CONSTRUCTING GRAPHS OF QUADRATIC EXPRESSIONS 

In each case, determine from the graph what values 
of x % if any, will make the expression equal to zero : 

1. ^-6*+ 8 5. x^-lx-Xb 

2. ^2_4 r + 3 6. ^ 2 -9 

3. ^2 + 7^+10 7. x 2 -8x + 16 

4. x 2 + x _ 12 a x * + 4:x+2 

Now that we have seen how to graph a quadratic 
EXPRESSION, we are ready to consider the first method 
of solving quadratic equations : 

B. HOW TO SOLVE QUADRATIC EQUATIONS 
I. GRAPHICAL SOLUTION 

Section 115. To solve a quadratic v equation is to find 
values of x which will make the expression equal to 
zero. In the previous section we graphed quadratic 
EXPRESSIONS. It is important to note that they were 
not quadratic EQUATIONS. Thus, ^r 2 -8^+12 is a 
quadratic expression, — that is, an algebraic expression of 
the second degree in one unknown, — but it is not an eqita- 



How to Solve Equations of the Second Degree 237 

Hon. If, however, we should write x 2 — 8x + 12 = 0, then 
we have a quadratic equation, — an equation of the second 
degree. To solve this equation is to find the value or 
values of x which will make x 2 — 8;r+12 equal to zero, 
because such values will make one member of the equation 
equal to the other ; i.e. = 0. Graphically, this can be 
done by plotting the values of the expression x 2 — 8x+ 12, 
as we did in the previous section. Thus, to graph the 
quadratic equation x 2 — 8.r + 12 = 0, we graph the ex- 
pression x 2 — 8 x + 12, and note from the graph what 
values of x will make 
the expression equal to 
zero. 

Thus, from the 
graph we see that 
x 2 — 8x + 12 is zero 
when x = 2 or when 
x = 6. Thus, we see 
that x can have two 
values in the quadratic 
equation 

^2 -8*+ 12=0. 

Checking, we find that 
2 and 6 each satisfies 
the equation. FlG m 

This suggests the 
following method for solving a quadratic equation graphi- 
cally : 

1. Graph the quadratic expression which forms one 
member of the equation. (The other member 
should be zero,) 















































1 












































































1 








, 




















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1 , 




























' r 


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238 Fundamentals of High School Mathematics 

'1. From the graph find for what values of x the 
expression is zero. In other words, find at ivJiat 
value of x tlie graph cuts the x-axis. These 
values are the values of x which will satisfy the 
equation. 

3. Check your result by substituting these values 
of the unknown in the original equation to see 
if they do satisfy it. 



EXERCISE 106 
PRACTICE IN SOLVING QUADRATIC EQUATIONS GRAPHICALLY 

Solve the following quadratic equations by drawing 
graphs of the expressions : 






1. 


x 2 - 5x - 14 = 


= 


6. 


2/+ 5/4-3 = 




2. 


A -2 + 3^=40 




7. 


x* + 8x + 16 = 




3. 


f - y = 20 




8. 


f + By + 1 = 




4. 


2x = 48 - x 2 




9. 


O- 2) 2 + 6;tr = 12 




5. 


x 2 -6x+9 = 





10. 


x 2 + 4 = 





In these examples, did you find any graph that did not 
cut the .r-axis in two places ? What conclusion would 
you draw if the graph just touched the ;r-axis ? if it did 
not even touch it ? 

What seem to be the disadvantages of the graphical 
method of solving quadratic equations/? the advantages ? 



How to Solve Equations of the Second Degree 239 

II. HOW TO SOLVE QUADRATIC EQUATIONS BY 
FACTORING THE EXPRESSION 

Section 116. By making use of a principle that you al- 
ready know, — namely, that the product of any number of 
factors is zero if one of them is zero, — we have a very easy 
method of solving quadratic equations. Recall that 
4.5.0 = 0, or that 6 • • 26 = or that a ■ b • c = if 
either a, or b, or c is 0. Why ? Because any number 
multiplied by is 0. Under what conditions is the prod- 
uct, xyz, zero ? Evidently, if eitlier x, y, or z is zero ; that 
is, if one of the factors is 0. In the same way, the expres- 
sion (x — 4)(;r+ 6), which is the product of two factors, 
could be if x— 4 were 0, or if x + 6 were 0. Now let 
us apply this to the solution of a quadratic equation. 

Illustrative example. The square of a certain number, in- 
creased by twice the number, gives as a result 48. What is the 
number ? 
Translating into algebraic language, we have the quadratic 

equation 

x 2 + 2* = 48. (1) 

In order to solve the equation we want to get a product equal to 
zero. Hence the equation x 2 -f 2 x = 48 should read, 

x 2 + 2* -48 = 0. (2) 

Factoring the left member to form a product, 

(x + 8)(x _ 6)=0 . (3) 

Note here that we have a product equal to 0. But, in order that a 
product can be 0, one of the factors must be 0. 

Therefore, if x + 8 = 0, x must equal - 8, or if x — 6 is to be 0, 

x must be 6. 

Summarizing the solution, we have the following steps : 

x 2 + 2jc=48. (1) 

Making one side 0, x 2 + 2 x — 48 = 0. (2) 

Forming a product, (x + 8)(jc - 6)= 0. (3) 

Making each factor 0, x + 8 = 0, or x = — 8. (4) 

x-6 = 0, orx=6. (5) 

Checking, by substituting 6 and — 8 in equation (1), 

64 - 16 = 48, 
or 36 + 12 = 48. 



240 Fundamentals of High School Mathematics 



EXERCISE 107 

Solve the following quadratic equations by factoring. 
Check each one. 

10. y + ^ = e 

2 
11. a 2 + a = 20 

13. r 2 +2^=0 

14. ^ 3 -2.r 2 = ^r 

15. 12 = ^ 2 + ^ 



16. The altitude of a triangle exceeds its base by 
4 inches ; the area is 96 square inches. Find 
the base and altitude. 

17. Find two consecutive even numbers whose 
product is 120. 

18. The sum of two numbers is 10 ; the sum of 
their squares is 52. Find each number. 



1. 


*a-5;r + 6 = 


2. 


^2_ J/ = 20 


3. 


* 2 + 9a = 22 


4. 


b 2 - 36 = 


5. 


2* 2 -5;r + 3 = 


6. 


e 2 - 8e= -16 


7. 


VI 1 =7/1 + 2 


8. 


fl 4- / = 56 


9. 


i^2 +r = 12 






Which of the two methods of solving a quadratic equa- 
tion that we have considered thus far do you think is 
better ? State the reasons for your choice. Could you 
solve any quadratic equation by the Factoring Method ? 
Solve x 1 + 3 x = 12 by this method. 



Before we can consider the third method of solving quadratic 
equations, — namely, solution by " completing the square," — it 
will be necessary to learn how to find the square root of algebraic 
expressions. 






How to Solve Equations of the Second Degree 241 

SQUARE ROOT 

Section 117. The need for square root. In arithmetic 
you learned how to find a side of a square when its area 
was known. For example, if you knew that the area of a 
square was 81 square inches, you learned to find one side 
of the square by finding one of the two equal factors of 
81, or, in other words, by finding the square root of 83 . 
Just now we cannot learn the best method of solving quad- 
ratic equations until we learn how to find the square root 
of algebraic expressions and of arithmetic numbers. 

Section 118. What is square root? The square root of 
a number is one of the two equal factors of that number. 
Let us illustrate with the number 100. We might factor 
it in several ways, as follows : 

4 x 25 = 100 

5 x 20 = 100 
10 x 10 = 100 

8 x 12.5 = 100, etc. 

Remembering that the square root of a number is one of 
its two equal factors, we see that neither 4 nor 25 is the 
square root of 100, because they are not equal factors. 
Evidently, 10 must be the square root. In the same way, 
— 10 is also a square root of 100 because it is one of two 
equal factors of 100. 

In the same way the cube root of a number is one of 
the tliree equal factors of the number. Thus, — 2 is the 
cube root of - 8 because (- 2)(- 2)(- 2) = - 8. 

Section 119. How to indicate the root of a number. It 
has been agreed to indicate the root of a number by the 
symbol V~ , called a radical sign. To designate what 



242 Fundamentals of High School Mathematics 

root is meant, a small number called an index, is placed 
in the radical sign. Thus, VlO means the fourth root of 
1(), i.e. 2, because 2 • 2 • - • - = 1(> ; V27 means the cube 
root of -7, i.e. 3, because 3 • 3 . 3 = 27 ; and V25 means 
the square root of 25. It is customary, however, to omit 
the index when square root is meant. Thus, V25, without 
an index, is always understood to mean the square root 
of 25. 



EXERCISE 108 

FIND, BY TRIAL, THE ROOTS,. WHICH ARE INDICATED, OF THE 
FOLLOWING EXPRESSIONS 






i. V9 io. V/ f 

2. V^« ,, J36.r 2 

3. vsra ' , »•>* 

4. *§ 12 ^^_ 

v 13. V144^ 10 

5. v 27 x* / <AA 9 . 

, i4. V4oo.f 2 y 

6 ' ^ la 15. ^125^? 

7. V64* 8 16 </i6^i 

8. VlOOy 5 i7. -v / 243y° 

9. V] 18. a/25G x 12 



Section 120. How to find the square root of algebraic 
expressions. We have seen that {a 4- bf or {a + b)(a + /;) 
= a 1 + 2 ab 4- b 2 . From this iV is evident that the square 
root of a 2 + 2 ab + b 2 must be a + b, or V# 2 + 2 ab + £ 2 
= (<? + b). In the same way Vx 2 + 10 x + 25 is x + 5, be- 
cause (/r + 5)(.# + 5) gives .r 2 + 1 x + 25. From these 
illustrations we see that it is possible to extract the square 



How to Solve Equations of the Second Degree 243 

root of an algebraic expression if we can show that it can 
be obtained by squaring some other expression ; that is, if 
we can show that it is the product of two equal factors. 

EXERCISE 109 

Find the square root of the following expressions, where 
it is possible to do so. Check each. 



1. 


x 2 + 2xy + y 2 


8. 


y 2 + 6y + 20 


2. 


;z 2 + 6rt + 9 


9. 


25^ 2 + 40^ + 16 


3. 


p _ 4 b + 4 


10. 


.r 2 + 16 


4. 


/ 2_ 10/+ 25 


11. 


y 2 ~ 49 


5. 


4<z 2 + 12tf + 9 


12. 


12^ + 36+^ 2 


6. 


1Cd+x 2 + 8x 


13. 


t 2 + u 2 + 2 /» 


7. 


1 + 21*+ 100 * 2 


14. 


r¥ - 6 n- 2 + 9 



If any of the expressions above are not perfect squares, 
make the necessary changes to transform them into perfect 
squares. 

HOW TO FIND THE SQUARE ROOT OF ARITHMETICAL 

NUMBERS 

Section 121. We have learned that the square root of a 
number is one of its two equal factors. Hence, to find the 
square root, we need to find one of its two equal factors. 

If a number is divided by its square root, the quotient 
will be the same as the divisor. Thus, if 36 is divided by 
its square root, 6, the quotient will be the same as the di- 
visor, 6. But if 36 is divided by a number which is smaller 
than its square root, the quotient will be larger than its 
square root. Thus, if 36 is divided by 4, the quotient is 9. 
The square root is somewhere between the divisor, 4, and 
the quotient, 9. If the divisor and quotient are the same, 



244 Fundamentals of High School Mat Hematics 

either one is the square root of the number, but if they are 
not the same, then the square root is some number between 
them. The next exercise will illustrate this trial method 
of finding square root. 

Find the square root of each of the following numbers 
by the trial method : 

First illustrative example. Find the square root of 55. 
Since this number is between the two perfect squares, 49 and 
64, its square root will be between 7 and 8 ; let us try 7.5 and 
use it as a divisor (one factor) to find the quotient (the other 
factor). 

7.333 



7.50)55.0 


52 5 


2 50 


2 25 


250 


225 



250 



This shows that the square root of 55 is between 7.500 and 7.333. 
Let us try a number halfway between them, say 7.416. 



7.4164 


7.4160)55.000 


51 912 


3 0880 


2 9664 


12160 


7416 


47440 


44496 


29540 


29664 






The square root, then, of 55 is 7.4162 This can be checked by 
multiplication or by division. 



How to Solve Equations of the Second Degree 245 

Second illustrative example. Find the square root of 12£. 
Since the number is between the two squares, 9 and 16, its 
square root will be between 3 and 4; let us try 3.5 and use it as 
a divisor. 

3.57 
3.5)12.5 
10 5 



2 00 
1 75 



250 

245 

5 

This shows that the square root of 12.5 is between 3.50 and 3.57. 
A value closer than either of these is the number halfway be- 
tween them, say 3.535. 

3.536 
3.535)12.5 

10 605 
1 8950 
1 7675 



12750 
10605 
21450 
21210 



This shows that the square root is between the two factors 
3.535 and 3.536. Using the number halfway between them, 
3.5355, we have the square root correct to 4 decimals. 



EXERCISE 110 

By using this method, find the square root of each of 
the following numbers : 

1. 18 4. 500 7. 965 10. 14.75 13. 3 

2. 52 5. 16.80 8. 3820 11. 2025 14. 2 

3. 200 6. 150 9. .2640 12. 8 15. 10 



246 Fundamentals of High School Mathematics 

Section 122. The old method of finding square root. 1 
Many pupils have learned in arithmetic another method of 
finding the square root of numbers. It is illustrated in the 
following example : 

Illustrative example. Find the square root of 200. 

Note the following steps : 200.0000)14.14 

(1) The number is separated into periods L 

of two figures each, counting from the 

decimal point. 24)100 

(2) You find the greatest square in the gg 
left-hand period, and write its square 9Q 1 /LAO 
root for the first figure of the root. ^ c ' 

(3) Subtract this square from the left- 281 
hand period, and with the remainder 11900 
place the next period for a new divi- 2824)11296 
dend. (This is 100 in the example.) chyL 

(4) Double the part of the root already 

found (2x1 = 2) for your trial divisor. Divide the dividend, 
exclusive of the right-hand figure (10) by the trial divisor, 2. 
Write the quotient obtained, 4, as the next figure of the root and 
the divisor. Multiply the complete divisor, 24, by the last 
term of the root, 4. Subtract the product, 96, from the dividend, 
100. To the remainder, 4, annex the next period, 00 for a new 
dividend. Repeat this process until all periods are used, or until 
any required degree of accuracy is obtained. 



EXERCISE 111 
Solve the examples of Exercise 110 by this method. 

1 It is believed that this traditional method is much more difficult to 
rationalize for the pupil than the so-called trial or estimate method. For 
those teachers who insist upon its use, the trial method may be omitted. 
In the interest of experimentation, however, the authors hope the proposed 
method will be fairly tested out. 






How to Solve Equations of the Second Degree 247 

III. QUADRATIC EQUATIONS SOLVED BY THE MOST 
GENERAL METHOD: COMPLETING THE SQUARE 

Section 123. The third method of solving quadratic equa- 
tions. Only a few easy quadratics can be solved by the 
second method which we studied, i.e. only those which can 
be factored. Furthermore, the first method shows that 
graphical solutions are too slow and often give only ap- 
proximate results. Consequently we need a more general 
method — one that is applicable to all quadratics, and one 
that gives accui'ate results. Let us illustrate such a general 
method of solving quadratic equations. 

First illustrative example. The area of a rectangle, which 
is 6 inches longer than it is wide, is 55 square inches. Find its 
dimensions. 

Translating into algebraic form, we have 

x 2 + 6x = 55. (1) 

Adding 9 to each side to make the left side a perfect square, 

x 2 + 6 x + 9 = 64. (2) 

Extracting the square root of each side, 

x + 3 = + 8 or - 8. (3) 

Using the +8, x = 5. (4) 

Using the -8, x=-li. (5) 

Checking the result : etc. 



Section 124. We must make the left side a perfect square. 
It is important to see why 9 was added to each side of 
equation (1). Why not add 10 or 20 or any other number, 
to each side ? Because tJie left side would not be a perfect 
square if any other number were added. Success with the 
method of completing the square depends upon knowing 
what number to add to each side to make the left side a 
perfect square. Note that the left side must be a perfect 
square, because you cannot extract the square root of a?i 
algebraic expression zvliich is not a perfect squa7'e. 



24^ Fundamentals of High School Mathematics 

Second illustrative example. Find a number such that 
its square decreased by 3 times itself shall be 10. 
Translating into algebraic form gives : 

x 2 -3;t = 10. (1) 

Adding $ to each side in order to make the left side a perfect 
square, 

X2_ 3x + } = 10 + £ = ^. (2) 

Extracting the square root of each side, 

x - f = | or - f. (3) 

Using + }, x = Af- or 5. (4) 

Using -I, x = - I or - 2. (5) 

Check : etc. 
.*. The unknown number is either -f- 5 or — 2. 



Here again, as in the first illustrative example, the most 
important and most difficult step is completing the sqziare, 
i.e. to know what to add to x* — 3 x to make it a perfect square. 
We need to know why |, rather than some other number, 
was added to each side. Remember that the algebraic 
expression, the left side of the equation, must be a perfect 
square, for otherwise we could not extract the square root 
of it. But we can extract the square root of the right side 
(because it is an arithmetical number) even if it is not a 
perfect square. 

Section 125. How to complete the square of any quadratic 
expression : add the square of one half the coefficient of x. 
Because we use the method so frequently, it will be worth 
while to learn the general method of completing the square. 

p 
If we square x + ^ , we get the expression 

that is, ( x + £) = x(2, +P x + ^t' 



How to Solve Equations of the Second Degree 249 

Note here that the last term of the expression is ^-- It 

P 
is the square of —' But p represents the coefficient of x 

in the expression which we wish to change into a perfect 
square. Thus, if we had the expression 

x 1 +$x 

and desired to change it into a perfect square, i.e. to com- 
plete the square, we should have to add — ; that is, add the 

square of one half the coefficient of x. 

To complete the square in the expression x 2 + 6x (see 
first illustrative example), we should add the square of one 
half the coefficient of x ; that is, the square of J of 6, or 9. 
In the same way, to complete the square in the expression 
x 2 — 3 x, we must add the square of one half the coefficient of 
x\ that is, the square of \ of 3 or (f) 2 , which is f. 

EXERCISE 112 

Solve each of the following quadratic equations by the 
method of completing the square : 

1. x 2 + 6x = ±Q 6. j 2 + 7j=8- 

2. * 2 -8;r=84 7. x 2 + x=56 

3. f + 2y=lb 8. ^ 2 +6^ = ll 

4. ^2_ 10 ^ = _ 1(3 9 p2_Ap = ^ 

5. /2 + 3/ = 10. 10. 10.r + ^ 2 =-9 

11. A rectangular field is 2 rods longer than it is 
wide, and it contains 6 acres. Find the length 
of the sides. 

12. Find two consecutive even numbers whose prod- 
uct is 80. 



250 Fundamentals of High School Mathematics 



13. Find the value of x in the equation 
Illustrative example. 

3x 2 -13x + 4 = 0. 

Solution : The equation must be in the form 

jc2 + />jc = n. 
In other words, the coefficient of x 2 must be 1. 
Dividing each side of (1) by 3 gives 

Adding (\ 3 ) 2 to each side gives 

v2_I3y J. 1_6 9 — 133. — 4. — 121 
* 3 x * 3^ — 3¥ 3 — 35 ' 

Extracting the square root of each side gives 

*-¥ = -¥ or -y. 

Using + V, x = ¥ + ¥ = ¥ = 4. 

Using --V-, x =¥-¥ = f = i 

Substituting J for x in equation (1), to check gives 
3.^-13.^ + 4 = 0. 
1 - ¥ + ¥ = o. 



(i) 



(2) 
(3) 
(4) 



The pupil should check for x ■ 

14. 2x 2 + 10 x=12 

15. 3a 2 + 6a = 45 



16. 



- + - = 9 
2 4 



Hint: Get rid of fractions. 

17. y-40 = 8y 



18. 



20. 



2;r 2 



3 
x 3 



— ;r= 3 



19. x — 



47 
2 * 10 

^,§£^81 

2 5 2 

21. 3j 2 + 5j = 22 

22. 5£ 2 +lG£+3: 







23. The difference of two numbers is 4, and the 
sum of their squares is 210. Find the number. 

24. A farmer has a square wheatfield containing 
10 acres. In harvesting the wheat, he cuts a 
strip of uniform width around the field. How 



How to Solve Equations of the Second Degree 251 

wide a strip must be cut in order to have the 
wheat half cut ? 

25. Divide 20 into two parts whose product is 96. 

26. The sum of two numbers is 20, and the sum of 
their squares is 208. Find the numbers. 

27. I went to the grocery for oranges. The clerk 
said they had advanced 10 cents per dozen. I 
got J dozen fewer oranges for a dollar. What 
was the original price per dozen? 

28. A piece of tin in the form of 
a square is taken to make an 
open-top box. The box is 
made by cutting out a 3-inch 
square from each corner of 
the piece of tin and folding 
up the sides. Find the length 
of the side of the original 
piece of tin if the box contains 243 cubic inches. 

29. A rectangular park 56 rods long and 16 rods 
wide is surrounded by a boulevard of uniform 
width. Find the width of this street if it con- 
tains 4 acres. 

30. The members of a high-school class agreed to 
pay $8 for a sleigh ride. As 4 were obliged to 
be absent, the cost for each of the rest was 10 
cents greater than it otherwise would have been. 
How many intended to go on the sleigh ride ? 

31. Solve by all three methods the following quad- 
ratic equation : 

2^ 2 + 5^ = 18. 




Fig. 112 



252 Fundamentals of High School Mathematics 

REVIEW EXERCISE 113 

1. The length of a 10-acre field is 4 times its 
width. What are its dimensions ? 

2. Does x = — I satisfy the equation 4 x 1 + 11 x= 
-6? 

3. The space passed over by a body falling t sec- 
onds is expressed by the formula 5 = 16/ 2 , 
where 5 is the number of feet the body falls. 
Construct a graph for this formula, using for t 
the values 0, J, 1, 11 2, 2J, and 3. Plot the 
values of / along the horizontal axis. 

4. Does V9 + Vl6 = V25? Does the sum of the 
square roots of two numbers equal the square 
root of the sum of the numbers ? Does 
Va + V6 = Va + 6? 

5. Is the square of a number always larger than 
the number ? Illustrate. 



6. Evaluate -\/s(s — a)(s — 6)(^ — c) if ^ = 20, a = 8, 
b = 14, and <: = 18. 

7. What is the square root of — 25 ? of — x 2 ? 

8. Does (x -f j/) 2 = .r 2 + jj/ 2 ? Test by using 4 for x 
and 3 forjj/. 

9. If you wanted to divide a product of several 
factors, such as 6 • 8 • 10 • 12, by some number 
such as 2, would you divide each of the factors 
by 2? 

10. Complete the statement: To divide a product 

by a number, divide of the factors by that 

number. 






How to Solve Equations of the Second Degree 253 

11. If a square piece of tin 10 inches on each side 
sells for 60 cents, what should a 15-inch square 
piece of the same thickness sell for ? 

12. A girl went to the bakery for pies. If she 
could buy pies 5 inches in diameter for 15 cents 
each, or 10 inches in diameter (the same thick- 
ness) for 35 cents, which would be cheaper for 
her to buy ? 

SUMMARY 

This chapter has taught : 

1. The meaning of a quadratic equation. 

2. How to solve quadratic equations by three 
methods : 

(a) By graphical representation. 

(6) By factoring. 

(c) By completing the square. 

3. Square root. 



CHAPTER XV 



FURTHER USE OF THE RIGHT TRIANGLE: HOW TO 
SOLVE QUADRATIC EQUATIONS WHICH CONTAIN 
TWO UNKNOWNS 

Section 126. Previous use of the right triangle. We 

have already seen how right triangles can be used to find 

unknown distances. If we knew one side, and one acute 

angle, we were able to find any other side. Now we come 

to another method of dealing with right triangles ; namely, 

when two sides are known, but when no acute angle is known. 

This method will be illustrated by the following problem : 

What is the longest straight 
line you can draw upon a rec- 
tangular blackboard 28 in. 
wide and 36 in. long ? 

Evidently the longest 
straight line is the diagonal 
of the blackboard, or the 
hypotenuse of the right tri- 
angle, Fig. 113. Thus, 
we need to know how to 
find the hypotenuse of a 
right triangle when the 
other two sides are known. 




Fig. 113 



This leads to the following : 



THE PYTHAGOREAN THEOREM 

Section 127. This important relation between the sides 
of a right triangle was discovered by the celebrated Greek 
mathematician Pythagoras, after whom it has been named. 
This theorem or law states that the square of the hypote- 
nuse of a right triangle is equal to the sum of the squares of 
the other two sides, or, in the above problem, that 

h* = 36 2 + 28 2 . 



254 



The Pythagorean Theorem 



255 



























A 









































Fig. m 



This relation between the sides of a right triangle can be 
seen from Fig. 114. The 
base, AB, and altitude, AC, 
of a right triangle are 
drawn so that they contain 
a common unit an integral 
number of times. AB con- 
tains the common unit 1 
times and AC contains it 
3 times. Then by actual 
measurement BC will con- 
tain the same unit 5 times. 
By constructing squares on 
the sides of the triangle, 
you can see by counting that 
the sum of the squares on AB and AC is equal to the 
square on BC. 

To test this further, the pupil should construct a right 
triangle with the base 12 units and the altitude 5 units. 
Then actually measure the hypotenuse, and note whether 
the square on the hypotenuse is equal to the sunt of the 
squares of the other two sides. Now we are ready to go 
back to the problem of finding the longest line that can be 
drawn upon the blackboard. By making use of the truth 
which was just studied we have : 
h 2 = 28 2 + 36 2 

or h 2 = 781 + 1296 

or h 2 = 2080 

or y£=V2080 = 45.66 in. 

This relation between the sides of a right triangle is 
more widely used by engineers, carpenters, mechanics, and 
builders than any other mathematical law. Historical 



'& 



Fundamentals of High School Mathematics 



records show that the knowledge of this important relation 
is as old as civilization itself. 

EXERCISE 114 

Problems based on the Pythagorean Theorem. 

1. A rectangular schoolroom- floor is 32 feet long 
and 28 feet wide. What is the longest straight 
line that could be drawn upon the floor ? 

2. How much walking is saved by cutting diago- 
nally across a rectangular plot of ground which 
is 25 rods wide and 42 rods long ? 

3. A tree 100 feet high was broken off by a storm. 
The top struck the ground 40 feet from the foot 
of the tree, the broken end remaining on the 
stump. Find the height of the part standing, as- 
suming the ground to be level. Make a drawing. 

4. What is the diagonal of a square whose sides 
are each 10 in. ? 

5. Find the side of a square whose diagonal is 20 
inches. 

6. Two vessels start from the same place, one sail- 
ing due northwest at the rate of 12 miles per 
hour, and the other sailing due southwest at 
the rate of 16 miles per hour. How far apart 
are they at the end of 3 hours ? 

7. The foot of a 36-foot ladder is 13 ft. 6 in. from the 
wall of a building against which the top is lean- 
ing. How high on the wall does the top reach ? 

8. A rope stretched from the top of a 62-foot pole 
just reaches the ground 16 feet from the foot 
of the pole. Find the length of the rope. 



The Pythagorean Theorem 



*5V 



9. The side of a square room is 21.5 feet. Find 
its diagonal correct to two decimals. 

10. What is the perimeter of a square whose diag- 
onal is 12 inches ? 

11. The side of a square is a. What represents its 
area ? its perimeter ? its diagonal ? 

12. A rectangle is four times as long as it is wide. 
Find its diagonal if its area is 576 square inches. 

13. Figure 115 is an equi- 
lateral triangle, and CD 
is perpendicular to AB. 
Find CD if each side 
of the triangle is 20 
inches. Then find the 
area of triangle ABC 

14. The area of a right 
triangle is 21 square 
inches. Its base is 6 inches, 
and hypotenuse. 

15. The diagonal of a square is d. Show that s 
(side) is — -• 

16. CD, the altitude of 
equilateral triangle 
ABC, is 16 inches. 
Find the sides of the 
triangle and its area. 

17. How long an umbrella 
will lie flat down on the 
bottom of a trunk whose inside dimensions are 
27 inches by 39 inches ? 




Find its altitude 





f\ 


1 1 






\ r1 




Di 


\B \f 


k- 


? — 


=» 



Fig. 116 






258 Fundamentals of High School Mathematics 

18. Can a circular wheel 8 feet in diameter be taken 
into a shop if the shop door is 4| feet wide and 
6 1 feet high? 

19. If A represents the area of a square, what will 
represent its perimeter ? its diagonal ? 

20. The sides of a triangle are 12, 16, and 24 inches. 
Is it a right triangle ? Why ? 

21. If you know two sides of a triangle, can you 
always find its area? Explain. 

22. The hypotenuse of a right triangle is 10 feet, 
and one of its sides is 2 feet longer than the 
other. Find the length of the sides. 

23. Find the area of a square whose diagonal is 12 
inches longer than one of its sides. 

24. Will an umbrella 30 inches long lie flat down in 
a suit case whose inside dimensions are 18 by 
25 inches ? 

25. A rectangle is 12 by 18 inches. How much 
must be added to its length to increase its diag- 
onal 4 inches ? 

26. In a right triangle one side is one unit less than 
twice the other side. The hypotenuse is 17 
units. What is the area of the triangle ? 

27. One side of a right triangle is 3 times as long 
as the other. The hypotenuse contains 30 
inches. Find the area of the triangle. 

28. The dimensions of a certain rectangular black- 
board, and the longest line which can be drawn 
upon it, are represented in feet by three con- 
secutive even numbers. Find the dimensions 
of the blackboard. 



Solving Quadratic Equations with Two Unknowns 259 

SOLVING QUADRATIC EQUATIONS WITH TWO UNKNOWNS 

Section 128. In the last chapter we solved quadratic 

equations in one unknown. Many quadratic equations, 

however, contain two, or more, unknowns. For example, 

consider the equation 

y = A 

which states that one number is equal to the square of 
another number. This equation contains two unknowns, x 
and j/, and is at the same time a second-degree or quad- 
ratic equation. A graph will help to show the relation 
between the variables or unknowns in this equation. 

Tabulating : 

Table 16 



If x is 


o 


1 


2 


3 


4 


5 


-1 


-2 


-3 


-4 


then y is 


o 


1 


4 


S 


16 


25 


1 


4 


3 


16 



These values of x and y are plotted in Fig. 117. It is 
evident both from the graph and from the table that there 











\ _L 






t i 






i t 






-i i_ 






n r 






U ~i 






r^ ^ 






: t ~f -, 












"i /~ 






\ i 












: i^^i 






±z 






_/\ 






^r . 






















































Y 





Fig. 117. The line shows relationship between two variables, when 
one equals the square of the other. 



260 Fundamentals of High School Mathematics 

is an indefinite number of sets of values of x and y wliich 

will satisfy the equation. Note that values of x are plotted 
along the horizontal axis, and values of y along the vertical 
axis. 

The graph of this equation may be thought of as 
answering the question which is suggested by the fol- 
lowing : 

What numbers are so related that one of them is equal to the 
square of the other ? 



EXERCISE 115 
Graph each of the following quadratic equations : 



1. 


y = x 2, + 4 X 


5. 


x = y 2 -f- 6 y 


2. 


y = x 2 — 4:X 


6. 


x={y- 3)2 


3. 


y=(x+2) 2 


7. 


xy = 60 


4. 


x = y 2 


8. 


y = x 2 — x — 


9. 


Illustrative exan 


lple. ( 


jraph the equa 



states that the sum of the squares of two numbers is 16. 

x 2 +y 2 = 16. 
Solving the equation for x gives 



16 



or 
Tabulating : 



x=±V16-i/2. 



Table 17 



Ifyis 





1 


2 


3' 


4 


5 


-1 


-2 


-3 


-4 


-5 




thenx is 


±4. 


±3.8 


*3.4 


±2.6 





* 


*3.8 


±34 


±2.6 





* 


* If y is larger 1 
becomes a nega 
ber. We call it 


han A, 

tive nu 

imagi 


then y 2 

Tiber. 

nary. 


is gre; 
But w< 


iter ths 
i canno 


n 16 
t ext 


and t 
-act tl 


he exp 
le squa 


ression 
re root 


under 
of a neg 


the r 
;ative 


adica 
num- 





Solving Quadratic Equations with Two Unknowns 261 



The graph of this equation is a circle whose radius is 4 
units, as in Fig. 118. 

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Fig. 118 
10. X 2 + y 2 = 25 11. f + x 1 = 40 

GRAPHICAL SOLUTION OF A PAIR OF EQUATIONS 

Section 129. In solving equations there must always be 
as many equations as there are unknown quantities. The 

examples which you have just graphed were quadratic 
equations in two unknowns, or two variables. There are 
many sets of values of the unknowns which will satisfy 
any one of these equations, just as there were many sets 
of values which would satisfy a first-degree equation in 
two variables, such as x+j/ = 10. To obtain a single set 
of values, or a limited number of sets Rvalues which satisfy 
a quadratic equation in two unknowns, we must have two 
equations. (There must always be as many different equa- 
tions as there are unknowns.) 



262 Fundamentals of High School Mathematics 

We shall now consider two equations : 
Illustrative example. 

(y 2 = 4x + 4. (1) 

I x + y = 2. (2) 

What set of values of x and y will satisfy both of these equations ? 

Let us first solve them graphically. 

Tabulating equation (1) 

1/2 = 4* + 4. 

Table 18 



If xis 





2 


3 


8 


-1 


-2 


then y is 


±2. 


±3.4 


*4. 


±6 





impossible 



Plotting these' points gives the curve shown in Fig. 119. 

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Y 

Flo. 119. The x -distance and the ^-distance of 
the points of intersection of the two lines give 
the value of x and of y which satisfy the two 
equations. 



•X 






Solving Quadratic Equations with Two Unknowns 263 

Plotting x -f y = 2 as in Chapter 11 gives the straight line. 
From Fig. 119 we see that the graphs intersect in two points, 
(0, 2) and (+8, — 6). Hence, there are two sets of values, and 
only two, which satisfy both equations : 

/* = • and (*=+ 8 
\ y = 2 \y=-Q 

Show by checking that these values do satisfy both equations. 



The important step in solving equations graphically is to 
determine the intersection points of their graphs. If their 
graphs intersect in only one point, then there is only one 
set of values of the unknowns which will satisfy both 
equations. There will be as many solutions as there are 
intersection points of their graphs. 



EXERCISE 116 

Solve graphically the following pairs of equations : 
1. 



2. 



;tr 2 = 4jj/ lx 2 + y 2 = 16 

2x+y = 12 3 ' [^+^ = 11.31 

y 2 + 2=x \y = x 2 — 4:X 

;r-3j/ = 4 * \2x-y = 5 



ALGEBRAIC SOLUTION 

Section 130. The algebraic solution is much easier than 
the graphic solution. To illustrate, take the first example 
in the previous exercise : 



264 Fundamentals of High School Mathematics 

Illustrative example. 

f* 2 = 4y, (i) 

\2x + y = 12. (2) 

From (2), y = 12-2x. (3) 

Substituting 12 — 2 x for y in (1) gives 

x 2 = 4(12 + 2*), (4) 

or X 2 = 4 8 _8x. (5) 

Solving by factoring, 

x 2 + 8* -48 = 0. • (6) 

(x+12)(x-4)=0. 

x = - 12 or + 4. 
If jc = — 12, thenz/ = 36. 

If x = 4, then y = 4. 

The pupil should check each pair of values. 



EXERCISE 117 



Solve by the algebraic method each of the following 
pairs of equations. Check each. 

x + j/ = 6 5 * \j/=2x + l 



[2ab = - ! 



x*+2x=y 6 . -— - 

2x+j>--= 12 [a-6 = 5 

x+y = 2 



\*^f = V 7 - W = -15 



4. 



h+^ = 6 



*£ = 5 1 2^r 2 — j/ 2 = — 9 



9. Find two numbers whose difference is 9 and the 

sum of whose squares is 221. 
10. The area of a rectangular field is 216 square 
rods, and its perimeter is 60 rods. What are 
its dimensions ? 



Solving Quadratic Equations with Two Unknowns 265 

11. The sum of two numbers is \ 9 - and their product 
is |. Find the numbers. 

12. The hypotenuse of a right triangle is 25 feet. 
Find the other two sides if you know that their 
sum is 35 feet. 

13. A piece of wire 30 inches long is bent into the 
form of a right triangle whose hypotenuse is 13 
inches. Find the other sides of the triangle. 

14. The area and the perimeter of a rectangle are 
each 25. What are its dimensions ? 

15. A photograph, 8 inches by 10 inches, is enlarged 
until it covers twice the original area, keeping 
the ratio of the length to the width unchanged. 
Find the sides of the enlarged photograph. 

16. In placing telephone poles between two places, 
it was found that if the poles were set 10 feet 
farther apart than originally planned, 4 poles 
fewer per mile were needed. How far apart 
were the poles placed at first ? 

REVIEW EXERCISE 118 

1. By substituting any value for x in x 2 — 1, 2jr, 
and x 2 + 1, show that the three numbers which 
result are sides of a right triangle. 

2. If the sides of a right triangle are 6 inches and 

8 inches, then the hypotenuse must be I 

inches. 

3. How can you tell when a triangle is a right tri- 
angle without measuring its angle ? Is the 
triangle whose sides are 5, 12, and 13 a right 
triangle ? Why ? 






266 Fundamentals of High School Mathematics 

4. What is the area of an equilateral triangle each 
of whose sides is 30 inches? 

5. How would you find the side of a square which 
had the same area as a circle with a radius of 12 
inches ? 

6. Write a formula for b if a, b, and c are the alti- 
tude, base, and hypotenuse of a right triangle ; 
similarly, a formula for a. 

7. Express the hypotenuse of a right triangle whose 
altitude exceeds its base by 6 inches. 

■ £ . \x + y = 12, 

8. Solve the pair of equations: \ 2 ~ _ ^ q 

9. When is it impossible to find the square root of 
a number ? 



A full and complete index will be supplied 

for the regular edition of this book. It 

has been decided to forego an index in the 

present Experimental Edition. 



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